function sayHi() {
console.log(name);
console.log(age);
var name = 'Lydia';
let age = 21;
}
sayHi();
- A:
Lydia
andundefined
- B:
Lydia
andReferenceError
- C:
ReferenceError
and21
- D:
undefined
andReferenceError
Answer
Within the function, we first declare the name
variable with the var
keyword. This means that the variable gets hoisted (memory space is set up during the creation phase) with the default value of undefined
, until we actually get to the line where we define the variable. We haven't defined the variable yet on the line where we try to log the name
variable, so it still holds the value of undefined
.
Variables with the let
keyword (and const
) are hoisted, but unlike var
, don't get initialized. They are not accessible before the line we declare (initialize) them. This is called the "temporal dead zone". When we try to access the variables before they are declared, JavaScript throws a ReferenceError
.
for (var i = 0; i < 3; i++) {
setTimeout(() => console.log(i), 1);
}
for (let i = 0; i < 3; i++) {
setTimeout(() => console.log(i), 1);
}
- A:
0 1 2
and0 1 2
- B:
0 1 2
and3 3 3
- C:
3 3 3
and0 1 2
Answer
Because of the event queue in JavaScript, the setTimeout
callback function is called after the loop has been executed. Since the variable i
in the first loop was declared using the var
keyword, this value was global. During the loop, we incremented the value of i
by 1
each time, using the unary operator ++
. By the time the setTimeout
callback function was invoked, i
was equal to 3
in the first example.
In the second loop, the variable i
was declared using the let
keyword: variables declared with the let
(and const
) keyword are block-scoped (a block is anything between { }
). During each iteration, i
will have a new value, and each value is scoped inside the loop.
const shape = {
radius: 10,
diameter() {
return this.radius * 2;
},
perimeter: () => 2 * Math.PI * this.radius,
};
console.log(shape.diameter());
console.log(shape.perimeter());
- A:
20
and62.83185307179586
- B:
20
andNaN
- C:
20
and63
- D:
NaN
and63
Answer
Note that the value of diameter
is a regular function, whereas the value of perimeter
is an arrow function.
With arrow functions, the this
keyword refers to its current surrounding scope, unlike regular functions! This means that when we call perimeter
, it doesn't refer to the shape object, but to its surrounding scope (window for example).
There is no value radius
on that object, which returns undefined
.
+true;
!'Lydia';
- A:
1
andfalse
- B:
false
andNaN
- C:
false
andfalse
Answer
The unary plus tries to convert an operand to a number. true
is 1
, and false
is 0
.
The string 'Lydia'
is a truthy value. What we're actually asking, is "is this truthy value falsy?". This returns false
.
const bird = {
size: 'small',
};
const mouse = {
name: 'Mickey',
small: true,
};
- A:
mouse.bird.size
is not valid - B:
mouse[bird.size]
is not valid - C:
mouse[bird["size"]]
is not valid - D: All of them are valid
Answer
In JavaScript, all object keys are strings (unless it's a Symbol). Even though we might not type them as strings, they are always converted into strings under the hood.
JavaScript interprets (or unboxes) statements. When we use bracket notation, it sees the first opening bracket [
and keeps going until it finds the closing bracket ]
. Only then, it will evaluate the statement.
mouse[bird.size]
: First it evaluates bird.size
, which is "small"
. mouse["small"]
returns true
However, with dot notation, this doesn't happen. mouse
does not have a key called bird
, which means that mouse.bird
is undefined
. Then, we ask for the size
using dot notation: mouse.bird.size
. Since mouse.bird
is undefined
, we're actually asking undefined.size
. This isn't valid, and will throw an error similar to Cannot read property "size" of undefined
.
let c = { greeting: 'Hey!' };
let d;
d = c;
c.greeting = 'Hello';
console.log(d.greeting);
- A:
Hello
- B:
Hey!
- C:
undefined
- D:
ReferenceError
- E:
TypeError
Answer
In JavaScript, all objects interact by reference when setting them equal to each other.
First, variable c
holds a value to an object. Later, we assign d
with the same reference that c
has to the object.
When you change one object, you change all of them.
let a = 3;
let b = new Number(3);
let c = 3;
console.log(a == b);
console.log(a === b);
console.log(b === c);
- A:
true
false
true
- B:
false
false
true
- C:
true
false
false
- D:
false
true
true
Answer
new Number()
is a built-in function constructor. Although it looks like a number, it's not really a number: it has a bunch of extra features and is an object.
When we use the ==
operator, it only checks whether it has the same value. They both have the value of 3
, so it returns true
.
However, when we use the ===
operator, both value and type should be the same. It's not: new Number()
is not a number, it's an object. Both return false.
class Chameleon {
static colorChange(newColor) {
this.newColor = newColor;
return this.newColor;
}
constructor({ newColor = 'green' } = {}) {
this.newColor = newColor;
}
}
const freddie = new Chameleon({ newColor: 'purple' });
console.log(freddie.colorChange('orange'));
- A:
orange
- B:
purple
- C:
green
- D:
TypeError
Answer
The colorChange
function is static. Static methods are designed to live only on the constructor in which they are created, and cannot be passed down to any children. Since freddie
is a child, the function is not passed down, and not available on the freddie
instance: a TypeError
is thrown.
let greeting;
greetign = {}; // Typo!
console.log(greetign);
- A:
{}
- B:
ReferenceError: greetign is not defined
- C:
undefined
Answer
It logs the object, because we just created an empty object on the global object! When we mistyped greeting
as greetign
, the JS interpreter actually saw this as global.greetign = {}
(or window.greetign = {}
in a browser).
In order to avoid this, we can use "use strict"
. This makes sure that you have declared a variable before setting it equal to anything.
function bark() {
console.log('Woof!');
}
bark.animal = 'dog';
- A: Nothing, this is totally fine!
- B:
SyntaxError
. You cannot add properties to a function this way. - C:
"Woof"
gets logged. - D:
ReferenceError
Answer
This is possible in JavaScript, because functions are objects! (Everything besides primitive types are objects)
A function is a special type of object. The code you write yourself isn't the actual function. The function is an object with properties. This property is invocable.
function Person(firstName, lastName) {
this.firstName = firstName;
this.lastName = lastName;
}
const member = new Person('Lydia', 'Hallie');
Person.getFullName = function() {
return `${this.firstName} ${this.lastName}`;
};
console.log(member.getFullName());
- A:
TypeError
- B:
SyntaxError
- C:
Lydia Hallie
- D:
undefined
undefined
Answer
You can't add properties to a constructor like you can with regular objects. If you want to add a feature to all objects at once, you have to use the prototype instead. So in this case,
Person.prototype.getFullName = function() {
return `${this.firstName} ${this.lastName}`;
};
would have made member.getFullName()
work. Why is this beneficial? Say that we added this method to the constructor itself. Maybe not every Person
instance needed this method. This would waste a lot of memory space, since they would still have that property, which takes of memory space for each instance. Instead, if we only add it to the prototype, we just have it at one spot in memory, yet they all have access to it!
function Person(firstName, lastName) {
this.firstName = firstName;
this.lastName = lastName;
}
const lydia = new Person('Lydia', 'Hallie');
const sarah = Person('Sarah', 'Smith');
console.log(lydia);
console.log(sarah);
- A:
Person {firstName: "Lydia", lastName: "Hallie"}
andundefined
- B:
Person {firstName: "Lydia", lastName: "Hallie"}
andPerson {firstName: "Sarah", lastName: "Smith"}
- C:
Person {firstName: "Lydia", lastName: "Hallie"}
and{}
- D:
Person {firstName: "Lydia", lastName: "Hallie"}
andReferenceError
Answer
For sarah
, we didn't use the new
keyword. When using new
, it refers to the new empty object we create. However, if you don't add new
it refers to the global object!
We said that this.firstName
equals "Sarah"
and this.lastName
equals "Smith"
. What we actually did, is defining global.firstName = 'Sarah'
and global.lastName = 'Smith'
. sarah
itself is left undefined
, since we don't return a value from the Person
function.
- A: Target > Capturing > Bubbling
- B: Bubbling > Target > Capturing
- C: Target > Bubbling > Capturing
- D: Capturing > Target > Bubbling
Answer
During the capturing phase, the event goes through the ancestor elements down to the target element. It then reaches the target element, and bubbling begins.
- A: true
- B: false
Answer
All objects have prototypes, except for the base object. The base object is the object created by the user, or an object that is created using the new
keyword. The base object has access to some methods and properties, such as .toString
. This is the reason why you can use built-in JavaScript methods! All of such methods are available on the prototype. Although JavaScript can't find it directly on your object, it goes down the prototype chain and finds it there, which makes it accessible for you.
function sum(a, b) {
return a + b;
}
sum(1, '2');
- A:
NaN
- B:
TypeError
- C:
"12"
- D:
3
Answer
JavaScript is a dynamically typed language: we don't specify what types certain variables are. Values can automatically be converted into another type without you knowing, which is called implicit type coercion. Coercion is converting from one type into another.
In this example, JavaScript converts the number 1
into a string, in order for the function to make sense and return a value. During the addition of a numeric type (1
) and a string type ('2'
), the number is treated as a string. We can concatenate strings like "Hello" + "World"
, so what's happening here is "1" + "2"
which returns "12"
.
let number = 0;
console.log(number++);
console.log(++number);
console.log(number);
- A:
1
1
2
- B:
1
2
2
- C:
0
2
2
- D:
0
1
2
Answer
The postfix unary operator ++
:
- Returns the value (this returns
0
) - Increments the value (number is now
1
)
The prefix unary operator ++
:
- Increments the value (number is now
2
) - Returns the value (this returns
2
)
This returns 0 2 2
.
function getPersonInfo(one, two, three) {
console.log(one);
console.log(two);
console.log(three);
}
const person = 'Lydia';
const age = 21;
getPersonInfo`${person} is ${age} years old`;
- A:
"Lydia"
21
["", " is ", " years old"]
- B:
["", " is ", " years old"]
"Lydia"
21
- C:
"Lydia"
["", " is ", " years old"]
21
Answer
If you use tagged template literals, the value of the first argument is always an array of the string values. The remaining arguments get the values of the passed expressions!
function checkAge(data) {
if (data === { age: 18 }) {
console.log('You are an adult!');
} else if (data == { age: 18 }) {
console.log('You are still an adult.');
} else {
console.log(`Hmm.. You don't have an age I guess`);
}
}
checkAge({ age: 18 });
- A:
You are an adult!
- B:
You are still an adult.
- C:
Hmm.. You don't have an age I guess
Answer
When testing equality, primitives are compared by their value, while objects are compared by their reference. JavaScript checks if the objects have a reference to the same location in memory.
The two objects that we are comparing don't have that: the object we passed as a parameter refers to a different location in memory than the object we used in order to check equality.
This is why both { age: 18 } === { age: 18 }
and { age: 18 } == { age: 18 }
return false
.
function getAge(...args) {
console.log(typeof args);
}
getAge(21);
- A:
"number"
- B:
"array"
- C:
"object"
- D:
"NaN"
Answer
The rest parameter (...args
.) lets us "collect" all remaining arguments into an array. An array is an object, so typeof args
returns "object"
function getAge() {
'use strict';
age = 21;
console.log(age);
}
getAge();
- A:
21
- B:
undefined
- C:
ReferenceError
- D:
TypeError
Answer
With "use strict"
, you can make sure that you don't accidentally declare global variables. We never declared the variable age
, and since we use "use strict"
, it will throw a reference error. If we didn't use "use strict"
, it would have worked, since the property age
would have gotten added to the global object.
const sum = eval('10*10+5');
- A:
105
- B:
"105"
- C:
TypeError
- D:
"10*10+5"
Answer
eval
evaluates codes that's passed as a string. If it's an expression, like in this case, it evaluates the expression. The expression is 10 * 10 + 5
. This returns the number 105
.
sessionStorage.setItem('cool_secret', 123);
- A: Forever, the data doesn't get lost.
- B: When the user closes the tab.
- C: When the user closes the entire browser, not only the tab.
- D: When the user shuts off their computer.
Answer
The data stored in sessionStorage
is removed after closing the tab.
If you used localStorage
, the data would've been there forever, unless for example localStorage.clear()
is invoked.
var num = 8;
var num = 10;
console.log(num);
- A:
8
- B:
10
- C:
SyntaxError
- D:
ReferenceError
Answer
With the var
keyword, you can declare multiple variables with the same name. The variable will then hold the latest value.
You cannot do this with let
or const
since they're block-scoped.
const obj = { 1: 'a', 2: 'b', 3: 'c' };
const set = new Set([1, 2, 3, 4, 5]);
obj.hasOwnProperty('1');
obj.hasOwnProperty(1);
set.has('1');
set.has(1);
- A:
false
true
false
true
- B:
false
true
true
true
- C:
true
true
false
true
- D:
true
true
true
true
Answer
All object keys (excluding Symbols) are strings under the hood, even if you don't type it yourself as a string. This is why obj.hasOwnProperty('1')
also returns true.
It doesn't work that way for a set. There is no '1'
in our set: set.has('1')
returns false
. It has the numeric type 1
, set.has(1)
returns true
.
const obj = { a: 'one', b: 'two', a: 'three' };
console.log(obj);
- A:
{ a: "one", b: "two" }
- B:
{ b: "two", a: "three" }
- C:
{ a: "three", b: "two" }
- D:
SyntaxError
Answer
If you have two keys with the same name, the key will be replaced. It will still be in its first position, but with the last specified value.
26. The JavaScript global execution context creates two things for you: the global object, and the "this" keyword.
- A: true
- B: false
- C: it depends
Answer
The base execution context is the global execution context: it's what's accessible everywhere in your code.
for (let i = 1; i < 5; i++) {
if (i === 3) continue;
console.log(i);
}
- A:
1
2
- B:
1
2
3
- C:
1
2
4
- D:
1
3
4
String.prototype.giveLydiaPizza = () => {
return 'Just give Lydia pizza already!';
};
const name = 'Lydia';
name.giveLydiaPizza();
- A:
"Just give Lydia pizza already!"
- B:
TypeError: not a function
- C:
SyntaxError
- D:
undefined
Answer
String
is a built-in constructor, which we can add properties to. I just added a method to its prototype. Primitive strings are automatically converted into a string object, generated by the string prototype function. So, all strings (string objects) have access to that method!
const a = {};
const b = { key: 'b' };
const c = { key: 'c' };
a[b] = 123;
a[c] = 456;
console.log(a[b]);
- A:
123
- B:
456
- C:
undefined
- D:
ReferenceError
Answer
Object keys are automatically converted into strings. We are trying to set an object as a key to object a
, with the value of 123
.
However, when we stringify an object, it becomes "[object Object]"
. So what we are saying here, is that a["object Object"] = 123
. Then, we can try to do the same again. c
is another object that we are implicitly stringifying. So then, a["object Object"] = 456
.
Then, we log a[b]
, which is actually a["object Object"]
. We just set that to 456
, so it returns 456
.
const foo = () => console.log('First');
const bar = () => setTimeout(() => console.log('Second'));
const baz = () => console.log('Third');
bar();
foo();
baz();
- A:
First
Second
Third
- B:
First
Third
Second
- C:
Second
First
Third
- D:
Second
Third
First
Answer
We have a setTimeout
function and invoked it first. Yet, it was logged last.
This is because in browsers, we don't just have the runtime engine, we also have something called a WebAPI
. The WebAPI
gives us the setTimeout
function to start with, and for example the DOM.
After the callback is pushed to the WebAPI, the setTimeout
function itself (but not the callback!) is popped off the stack.
Now, foo
gets invoked, and "First"
is being logged.
foo
is popped off the stack, and baz
gets invoked. "Third"
gets logged.
The WebAPI can't just add stuff to the stack whenever it's ready. Instead, it pushes the callback function to something called the queue.
This is where an event loop starts to work. An event loop looks at the stack and task queue. If the stack is empty, it takes the first thing on the queue and pushes it onto the stack.
bar
gets invoked, "Second"
gets logged, and it's popped off the stack.
<div onclick="console.log('first div')">
<div onclick="console.log('second div')">
<button onclick="console.log('button')">
Click!
</button>
</div>
</div>
- A: Outer
div
- B: Inner
div
- C:
button
- D: An array of all nested elements.
Answer
The deepest nested element that caused the event is the target of the event. You can stop bubbling by event.stopPropagation
<div onclick="console.log('div')">
<p onclick="console.log('p')">
Click here!
</p>
</div>
- A:
p
div
- B:
div
p
- C:
p
- D:
div
Answer
If we click p
, we see two logs: p
and div
. During event propagation, there are 3 phases: capturing, target, and bubbling. By default, event handlers are executed in the bubbling phase (unless you set useCapture
to true
). It goes from the deepest nested element outwards.
const person = { name: 'Lydia' };
function sayHi(age) {
return `${this.name} is ${age}`;
}
console.log(sayHi.call(person, 21));
console.log(sayHi.bind(person, 21));
- A:
undefined is 21
Lydia is 21
- B:
function
function
- C:
Lydia is 21
Lydia is 21
- D:
Lydia is 21
function
Answer
With both, we can pass the object to which we want the this
keyword to refer to. However, .call
is also executed immediately!
.bind.
returns a copy of the function, but with a bound context! It is not executed immediately.
function sayHi() {
return (() => 0)();
}
console.log(typeof sayHi());
- A:
"object"
- B:
"number"
- C:
"function"
- D:
"undefined"
Answer
The sayHi
function returns the returned value of the immediately invoked function (IIFE). This function returned 0
, which is type "number"
.
FYI: there are only 7 built-in types: null
, undefined
, boolean
, number
, string
, object
, and symbol
. "function"
is not a type, since functions are objects, it's of type "object"
.
0;
new Number(0);
('');
(' ');
new Boolean(false);
undefined;
- A:
0
,''
,undefined
- B:
0
,new Number(0)
,''
,new Boolean(false)
,undefined
- C:
0
,''
,new Boolean(false)
,undefined
- D: All of them are falsy
Answer
There are only six falsy values:
undefined
null
NaN
0
''
(empty string)false
Function constructors, like new Number
and new Boolean
are truthy.
console.log(typeof typeof 1);
- A:
"number"
- B:
"string"
- C:
"object"
- D:
"undefined"
const numbers = [1, 2, 3];
numbers[10] = 11;
console.log(numbers);
- A:
[1, 2, 3, 7 x null, 11]
- B:
[1, 2, 3, 11]
- C:
[1, 2, 3, 7 x empty, 11]
- D:
SyntaxError
Answer
When you set a value to an element in an array that exceeds the length of the array, JavaScript creates something called "empty slots". These actually have the value of undefined
, but you will see something like:
[1, 2, 3, 7 x empty, 11]
depending on where you run it (it's different for every browser, node, etc.)
(() => {
let x, y;
try {
throw new Error();
} catch (x) {
(x = 1), (y = 2);
console.log(x);
}
console.log(x);
console.log(y);
})();
- A:
1
undefined
2
- B:
undefined
undefined
undefined
- C:
1
1
2
- D:
1
undefined
undefined
Answer
The catch
block receives the argument x
. This is not the same x
as the variable when we pass arguments. This variable x
is block-scoped.
Later, we set this block-scoped variable equal to 1
, and set the value of the variable y
. Now, we log the block-scoped variable x
, which is equal to 1
.
Outside of the catch
block, x
is still undefined
, and y
is 2
. When we want to console.log(x)
outside of the catch
block, it returns undefined
, and y
returns 2
.
- A: primitive or object
- B: function or object
- C: trick question! only objects
- D: number or object
Answer
JavaScript only has primitive types and objects.
Primitive types are boolean
, null
, undefined
, bigint
, number
, string
, and symbol
.
What differentiates a primitive from an object is that primitives do not have any properties or methods; however, you'll note that 'foo'.toUpperCase()
evaluates to 'FOO'
and does not result in a TypeError
. This is because when you try to access a property or method on a primitive like a string, JavaScript will implicitly wrap the object using one of the wrapper classes, i.e. String
, and then immediately discard the wrapper after the expression evaluates. All primitives except for null
and undefined
exhibit this behaviour.
[[0, 1], [2, 3]].reduce(
(acc, cur) => {
return acc.concat(cur);
},
[1, 2],
);
- A:
[0, 1, 2, 3, 1, 2]
- B:
[6, 1, 2]
- C:
[1, 2, 0, 1, 2, 3]
- D:
[1, 2, 6]
Answer
[1, 2]
is our initial value. This is the value we start with, and the value of the very first acc
. During the first round, acc
is [1,2]
, and cur
is [0, 1]
. We concatenate them, which results in [1, 2, 0, 1]
.
Then, [1, 2, 0, 1]
is acc
and [2, 3]
is cur
. We concatenate them, and get [1, 2, 0, 1, 2, 3]
!!null;
!!'';
!!1;
- A:
false
true
false
- B:
false
false
true
- C:
false
true
true
- D:
true
true
false
Answer
null
is falsy. !null
returns true
. !true
returns false
.
""
is falsy. !""
returns true
. !true
returns false
.
1
is truthy. !1
returns false
. !false
returns true
.
setInterval(() => console.log('Hi'), 1000);
- A: a unique id
- B: the amount of milliseconds specified
- C: the passed function
- D:
undefined
Answer
It returns a unique id. This id can be used to clear that interval with the clearInterval()
function.
[...'Lydia'];
- A:
["L", "y", "d", "i", "a"]
- B:
["Lydia"]
- C:
[[], "Lydia"]
- D:
[["L", "y", "d", "i", "a"]]
Answer
A string is an iterable. The spread operator maps every character of an iterable to one element.
function* generator(i) {
yield i;
yield i * 2;
}
const gen = generator(10);
console.log(gen.next().value);
console.log(gen.next().value);
- A:
[0, 10], [10, 20]
- B:
20, 20
- C:
10, 20
- D:
0, 10 and 10, 20
Answer
Regular functions cannot be stopped mid-way after invocation. However, a generator function can be "stopped" midway, and later continue from where it stopped. Every time a generator function encounters a yield
keyword, the function yields the value specified after it. Note that the generator function in that case doesnβt return the value, it yields the value.
First, we initialize the generator function with i
equal to 10
. We invoke the generator function using the next()
method. The first time we invoke the generator function, i
is equal to 10
. It encounters the first yield
keyword: it yields the value of i
. The generator is now "paused", and 10
gets logged.
Then, we invoke the function again with the next()
method. It starts to continue where it stopped previously, still with i
equal to 10
. Now, it encounters the next yield
keyword, and yields i * 2
. i
is equal to 10
, so it returns 10 * 2
, which is 20
. This results in 10, 20
.
const firstPromise = new Promise((res, rej) => {
setTimeout(res, 500, 'one');
});
const secondPromise = new Promise((res, rej) => {
setTimeout(res, 100, 'two');
});
Promise.race([firstPromise, secondPromise]).then(res => console.log(res));
- A:
"one"
- B:
"two"
- C:
"two" "one"
- D:
"one" "two"
Answer
When we pass multiple promises to the Promise.race
method, it resolves/rejects the first promise that resolves/rejects. To the setTimeout
method, we pass a timer: 500ms for the first promise (firstPromise
), and 100ms for the second promise (secondPromise
). This means that the secondPromise
resolves first with the value of 'two'
. res
now holds the value of 'two'
, which gets logged.
let person = { name: 'Lydia' };
const members = [person];
person = null;
console.log(members);
- A:
null
- B:
[null]
- C:
[{}]
- D:
[{ name: "Lydia" }]
Answer
First, we declare a variable person
with the value of an object that has a name
property.
Then, we declare a variable called members
. We set the first element of that array equal to the value of the person
variable. Objects interact by reference when setting them equal to each other. When you assign a reference from one variable to another, you make a copy of that reference. (note that they don't have the same reference!)
Then, we set the variable person
equal to null
.
We are only modifying the value of the person
variable, and not the first element in the array, since that element has a different (copied) reference to the object. The first element in members
still holds its reference to the original object. When we log the members
array, the first element still holds the value of the object, which gets logged.
const person = {
name: 'Lydia',
age: 21,
};
for (const item in person) {
console.log(item);
}
- A:
{ name: "Lydia" }, { age: 21 }
- B:
"name", "age"
- C:
"Lydia", 21
- D:
["name", "Lydia"], ["age", 21]
Answer
With a for-in
loop, we can iterate through object keys, in this case name
and age
. Under the hood, object keys are strings (if they're not a Symbol). On every loop, we set the value of item
equal to the current key itβs iterating over. First, item
is equal to name
, and gets logged. Then, item
is equal to age
, which gets logged.
console.log(3 + 4 + '5');
- A:
"345"
- B:
"75"
- C:
12
- D:
"12"
Answer
Operator associativity is the order in which the compiler evaluates the expressions, either left-to-right or right-to-left. This only happens if all operators have the same precedence. We only have one type of operator: +
. For addition, the associativity is left-to-right.
3 + 4
gets evaluated first. This results in the number 7
.
7 + '5'
results in "75"
because of coercion. JavaScript converts the number 7
into a string, see question 15. We can concatenate two strings using the +
operator. "7" + "5"
results in "75"
.
const num = parseInt('7*6', 10);
- A:
42
- B:
"42"
- C:
7
- D:
NaN
Answer
Only the first numbers in the string is returned. Based on the radix (the second argument in order to specify what type of number we want to parse it to: base 10, hexadecimal, octal, binary, etc.), the parseInt
checks whether the characters in the string are valid. Once it encounters a character that isn't a valid number in the radix, it stops parsing and ignores the following characters.
*
is not a valid number. It only parses "7"
into the decimal 7
. num
now holds the value of 7
.
[1, 2, 3].map(num => {
if (typeof num === 'number') return;
return num * 2;
});
- A:
[]
- B:
[null, null, null]
- C:
[undefined, undefined, undefined]
- D:
[ 3 x empty ]
Answer
When mapping over the array, the value of num
is equal to the element itβs currently looping over. In this case, the elements are numbers, so the condition of the if statement typeof num === "number"
returns true
. The map function creates a new array and inserts the values returned from the function.
However, we donβt return a value. When we donβt return a value from the function, the function returns undefined
. For every element in the array, the function block gets called, so for each element we return undefined
.
function getInfo(member, year) {
member.name = 'Lydia';
year = '1998';
}
const person = { name: 'Sarah' };
const birthYear = '1997';
getInfo(person, birthYear);
console.log(person, birthYear);
- A:
{ name: "Lydia" }, "1997"
- B:
{ name: "Sarah" }, "1998"
- C:
{ name: "Lydia" }, "1998"
- D:
{ name: "Sarah" }, "1997"
Answer
Arguments are passed by value, unless their value is an object, then they're passed by reference. birthYear
is passed by value, since it's a string, not an object. When we pass arguments by value, a copy of that value is created (see question 46).
The variable birthYear
has a reference to the value "1997"
. The argument year
also has a reference to the value "1997"
, but it's not the same value as birthYear
has a reference to. When we update the value of year
by setting year
equal to "1998"
, we are only updating the value of year
. birthYear
is still equal to "1997"
.
The value of person
is an object. The argument member
has a (copied) reference to the same object. When we modify a property of the object member
has a reference to, the value of person
will also be modified, since they both have a reference to the same object. person
's name
property is now equal to the value "Lydia"
function greeting() {
throw 'Hello world!';
}
function sayHi() {
try {
const data = greeting();
console.log('It worked!', data);
} catch (e) {
console.log('Oh no an error:', e);
}
}
sayHi();
- A:
It worked! Hello world!
- B:
Oh no an error: undefined
- C:
SyntaxError: can only throw Error objects
- D:
Oh no an error: Hello world!
Answer
With the throw
statement, we can create custom errors. With this statement, you can throw exceptions. An exception can be a string, a number, a boolean or an object. In this case, our exception is the string 'Hello world'
.
With the catch
statement, we can specify what to do if an exception is thrown in the try
block. An exception is thrown: the string 'Hello world'
. e
is now equal to that string, which we log. This results in 'Oh an error: Hello world'
.
function Car() {
this.make = 'Lamborghini';
return { make: 'Maserati' };
}
const myCar = new Car();
console.log(myCar.make);
- A:
"Lamborghini"
- B:
"Maserati"
- C:
ReferenceError
- D:
TypeError
Answer
When you return a property, the value of the property is equal to the returned value, not the value set in the constructor function. We return the string "Maserati"
, so myCar.make
is equal to "Maserati"
.
(() => {
let x = (y = 10);
})();
console.log(typeof x);
console.log(typeof y);
- A:
"undefined", "number"
- B:
"number", "number"
- C:
"object", "number"
- D:
"number", "undefined"
Answer
let x = y = 10;
is actually shorthand for:
y = 10;
let x = y;
When we set y
equal to 10
, we actually add a property y
to the global object (window
in browser, global
in Node). In a browser, window.y
is now equal to 10
.
Then, we declare a variable x
with the value of y
, which is 10
. Variables declared with the let
keyword are block scoped, they are only defined within the block they're declared in; the immediately-invoked function (IIFE) in this case. When we use the typeof
operator, the operand x
is not defined: we are trying to access x
outside of the block it's declared in. This means that x
is not defined. Values who haven't been assigned a value or declared are of type "undefined"
. console.log(typeof x)
returns "undefined"
.
However, we created a global variable y
when setting y
equal to 10
. This value is accessible anywhere in our code. y
is defined, and holds a value of type "number"
. console.log(typeof y)
returns "number"
.
class Dog {
constructor(name) {
this.name = name;
}
}
Dog.prototype.bark = function() {
console.log(`Woof I am ${this.name}`);
};
const pet = new Dog('Mara');
pet.bark();
delete Dog.prototype.bark;
pet.bark();
- A:
"Woof I am Mara"
,TypeError
- B:
"Woof I am Mara"
,"Woof I am Mara"
- C:
"Woof I am Mara"
,undefined
- D:
TypeError
,TypeError
Answer
We can delete properties from objects using the delete
keyword, also on the prototype. By deleting a property on the prototype, it is not available anymore in the prototype chain. In this case, the bark
function is not available anymore on the prototype after delete Dog.prototype.bark
, yet we still try to access it.
When we try to invoke something that is not a function, a TypeError
is thrown. In this case TypeError: pet.bark is not a function
, since pet.bark
is undefined
.
const set = new Set([1, 1, 2, 3, 4]);
console.log(set);
- A:
[1, 1, 2, 3, 4]
- B:
[1, 2, 3, 4]
- C:
{1, 1, 2, 3, 4}
- D:
{1, 2, 3, 4}
Answer
The Set
object is a collection of unique values: a value can only occur once in a set.
We passed the iterable [1, 1, 2, 3, 4]
with a duplicate value 1
. Since we cannot have two of the same values in a set, one of them is removed. This results in {1, 2, 3, 4}
.
// counter.js
let counter = 10;
export default counter;
// index.js
import myCounter from './counter';
myCounter += 1;
console.log(myCounter);
- A:
10
- B:
11
- C:
Error
- D:
NaN
Answer
An imported module is read-only: you cannot modify the imported module. Only the module that exports them can change its value.
When we try to increment the value of myCounter
, it throws an error: myCounter
is read-only and cannot be modified.
const name = 'Lydia';
age = 21;
console.log(delete name);
console.log(delete age);
- A:
false
,true
- B:
"Lydia"
,21
- C:
true
,true
- D:
undefined
,undefined
Answer
The delete
operator returns a boolean value: true
on a successful deletion, else it'll return false
. However, variables declared with the var
, const
or let
keyword cannot be deleted using the delete
operator.
The name
variable was declared with a const
keyword, so its deletion is not successful: false
is returned. When we set age
equal to 21
, we actually added a property called age
to the global object. You can successfully delete properties from objects this way, also the global object, so delete age
returns true
.
const numbers = [1, 2, 3, 4, 5];
const [y] = numbers;
console.log(y);
- A:
[[1, 2, 3, 4, 5]]
- B:
[1, 2, 3, 4, 5]
- C:
1
- D:
[1]
Answer
We can unpack values from arrays or properties from objects through destructuring. For example:
[a, b] = [1, 2];
The value of a
is now 1
, and the value of b
is now 2
. What we actually did in the question, is:
[y] = [1, 2, 3, 4, 5];
This means that the value of y
is equal to the first value in the array, which is the number 1
. When we log y
, 1
is returned.
const user = { name: 'Lydia', age: 21 };
const admin = { admin: true, ...user };
console.log(admin);
- A:
{ admin: true, user: { name: "Lydia", age: 21 } }
- B:
{ admin: true, name: "Lydia", age: 21 }
- C:
{ admin: true, user: ["Lydia", 21] }
- D:
{ admin: true }
Answer
It's possible to combine objects using the spread operator ...
. It lets you create copies of the key/value pairs of one object, and add them to another object. In this case, we create copies of the user
object, and add them to the admin
object. The admin
object now contains the copied key/value pairs, which results in { admin: true, name: "Lydia", age: 21 }
.
const person = { name: 'Lydia' };
Object.defineProperty(person, 'age', { value: 21 });
console.log(person);
console.log(Object.keys(person));
- A:
{ name: "Lydia", age: 21 }
,["name", "age"]
- B:
{ name: "Lydia", age: 21 }
,["name"]
- C:
{ name: "Lydia"}
,["name", "age"]
- D:
{ name: "Lydia"}
,["age"]
Answer
With the defineProperty
method, we can add new properties to an object, or modify existing ones. When we add a property to an object using the defineProperty
method, they are by default not enumerable. The Object.keys
method returns all enumerable property names from an object, in this case only "name"
.
Properties added using the defineProperty
method are immutable by default. You can override this behavior using the writable
, configurable
and enumerable
properties. This way, the defineProperty
method gives you a lot more control over the properties you're adding to an object.
const settings = {
username: 'lydiahallie',
level: 19,
health: 90,
};
const data = JSON.stringify(settings, ['level', 'health']);
console.log(data);
- A:
"{"level":19, "health":90}"
- B:
"{"username": "lydiahallie"}"
- C:
"["level", "health"]"
- D:
"{"username": "lydiahallie", "level":19, "health":90}"
Answer
The second argument of JSON.stringify
is the replacer. The replacer can either be a function or an array, and lets you control what and how the values should be stringified.
If the replacer is an array, only the property names included in the array will be added to the JSON string. In this case, only the properties with the names "level"
and "health"
are included, "username"
is excluded. data
is now equal to "{"level":19, "health":90}"
.
If the replacer is a function, this function gets called on every property in the object you're stringifying. The value returned from this function will be the value of the property when it's added to the JSON string. If the value is undefined
, this property is excluded from the JSON string.
let num = 10;
const increaseNumber = () => num++;
const increasePassedNumber = number => number++;
const num1 = increaseNumber();
const num2 = increasePassedNumber(num1);
console.log(num1);
console.log(num2);
- A:
10
,10
- B:
10
,11
- C:
11
,11
- D:
11
,12
Answer
The unary operator ++
first returns the value of the operand, then increments the value of the operand. The value of num1
is 10
, since the increaseNumber
function first returns the value of num
, which is 10
, and only increments the value of num
afterwards.
num2
is 10
, since we passed num1
to the increasePassedNumber
. number
is equal to 10
(the value of num1
. Again, the unary operator ++
first returns the value of the operand, then increments the value of the operand. The value of number
is 10
, so num2
is equal to 10
.
const value = { number: 10 };
const multiply = (x = { ...value }) => {
console.log((x.number *= 2));
};
multiply();
multiply();
multiply(value);
multiply(value);
- A:
20
,40
,80
,160
- B:
20
,40
,20
,40
- C:
20
,20
,20
,40
- D:
NaN
,NaN
,20
,40
Answer
In ES6, we can initialize parameters with a default value. The value of the parameter will be the default value, if no other value has been passed to the function, or if the value of the parameter is "undefined"
. In this case, we spread the properties of the value
object into a new object, so x
has the default value of { number: 10 }
.
The default argument is evaluated at call time! Every time we call the function, a new object is created. We invoke the multiply
function the first two times without passing a value: x
has the default value of { number: 10 }
. We then log the multiplied value of that number, which is 20
.
The third time we invoke multiply, we do pass an argument: the object called value
. The *=
operator is actually shorthand for x.number = x.number * 2
: we modify the value of x.number
, and log the multiplied value 20
.
The fourth time, we pass the value
object again. x.number
was previously modified to 20
, so x.number *= 2
logs 40
.
[1, 2, 3, 4].reduce((x, y) => console.log(x, y));
- A:
1
2
and3
3
and6
4
- B:
1
2
and2
3
and3
4
- C:
1
undefined
and2
undefined
and3
undefined
and4
undefined
- D:
1
2
andundefined
3
andundefined
4
Answer
The first argument that the reduce
method receives is the accumulator, x
in this case. The second argument is the current value, y
. With the reduce method, we execute a callback function on every element in the array, which could ultimately result in one single value.
In this example, we are not returning any values, we are simply logging the values of the accumulator and the current value.
The value of the accumulator is equal to the previously returned value of the callback function. If you don't pass the optional initialValue
argument to the reduce
method, the accumulator is equal to the first element on the first call.
On the first call, the accumulator (x
) is 1
, and the current value (y
) is 2
. We don't return from the callback function, we log the accumulator and current value: 1
and 2
get logged.
If you don't return a value from a function, it returns undefined
. On the next call, the accumulator is undefined
, and the current value is 3
. undefined
and 3
get logged.
On the fourth call, we again don't return from the callback function. The accumulator is again undefined
, and the current value is 4
. undefined
and 4
get logged.
class Dog {
constructor(name) {
this.name = name;
}
};
class Labrador extends Dog {
// 1
constructor(name, size) {
this.size = size;
}
// 2
constructor(name, size) {
super(name);
this.size = size;
}
// 3
constructor(size) {
super(name);
this.size = size;
}
// 4
constructor(name, size) {
this.name = name;
this.size = size;
}
};
- A: 1
- B: 2
- C: 3
- D: 4
Answer
In a derived class, you cannot access the this
keyword before calling super
. If you try to do that, it will throw a ReferenceError: 1 and 4 would throw a reference error.
With the super
keyword, we call that parent class's constructor with the given arguments. The parent's constructor receives the name
argument, so we need to pass name
to super
.
The Labrador
class receives two arguments, name
since it extends Dog
, and size
as an extra property on the Labrador
class. They both need to be passed to the constructor function on Labrador
, which is done correctly using constructor 2.
// index.js
console.log('running index.js');
import { sum } from './sum.js';
console.log(sum(1, 2));
// sum.js
console.log('running sum.js');
export const sum = (a, b) => a + b;
- A:
running index.js
,running sum.js
,3
- B:
running sum.js
,running index.js
,3
- C:
running sum.js
,3
,running index.js
- D:
running index.js
,undefined
,running sum.js
Answer
With the import
keyword, all imported modules are pre-parsed. This means that the imported modules get run first, the code in the file which imports the module gets executed after.
This is a difference between require()
in CommonJS and import
! With require()
, you can load dependencies on demand while the code is being run. If we would have used require
instead of import
, running index.js
, running sum.js
, 3
would have been logged to the console.
console.log(Number(2) === Number(2));
console.log(Boolean(false) === Boolean(false));
console.log(Symbol('foo') === Symbol('foo'));
- A:
true
,true
,false
- B:
false
,true
,false
- C:
true
,false
,true
- D:
true
,true
,true
Answer
Every Symbol is entirely unique. The purpose of the argument passed to the Symbol is to give the Symbol a description. The value of the Symbol is not dependent on the passed argument. As we test equality, we are creating two entirely new symbols: the first Symbol('foo')
, and the second Symbol('foo')
. These two values are unique and not equal to each other, Symbol('foo') === Symbol('foo')
returns false
.
const name = 'Lydia Hallie';
console.log(name.padStart(13));
console.log(name.padStart(2));
- A:
"Lydia Hallie"
,"Lydia Hallie"
- B:
" Lydia Hallie"
," Lydia Hallie"
("[13x whitespace]Lydia Hallie"
,"[2x whitespace]Lydia Hallie"
) - C:
" Lydia Hallie"
,"Lydia Hallie"
("[1x whitespace]Lydia Hallie"
,"Lydia Hallie"
) - D:
"Lydia Hallie"
,"Lyd"
,
Answer
With the padStart
method, we can add padding to the beginning of a string. The value passed to this method is the total length of the string together with the padding. The string "Lydia Hallie"
has a length of 12
. name.padStart(13)
inserts 1 space at the start of the string, because 12 + 1 is 13.
If the argument passed to the padStart
method is smaller than the length of the array, no padding will be added.
console.log('π₯' + 'π»');
- A:
"π₯π»"
- B:
257548
- C: A string containing their code points
- D: Error
Answer
With the +
operator, you can concatenate strings. In this case, we are concatenating the string "π₯"
with the string "π»"
, resulting in "π₯π»"
.
function* startGame() {
const answer = yield 'Do you love JavaScript?';
if (answer !== 'Yes') {
return "Oh wow... Guess we're gone here";
}
return 'JavaScript loves you back β€οΈ';
}
const game = startGame();
console.log(/* 1 */); // Do you love JavaScript?
console.log(/* 2 */); // JavaScript loves you back β€οΈ
- A:
game.next("Yes").value
andgame.next().value
- B:
game.next.value("Yes")
andgame.next.value()
- C:
game.next().value
andgame.next("Yes").value
- D:
game.next.value()
andgame.next.value("Yes")
Answer
A generator function "pauses" its execution when it sees the yield
keyword. First, we have to let the function yield the string "Do you love JavaScript?", which can be done by calling game.next().value
.
Every line is executed, until it finds the first yield
keyword. There is a yield
keyword on the first line within the function: the execution stops with the first yield! This means that the variable answer
is not defined yet!
When we call game.next("Yes").value
, the previous yield
is replaced with the value of the parameters passed to the next()
function, "Yes"
in this case. The value of the variable answer
is now equal to "Yes"
. The condition of the if-statement returns false
, and JavaScript loves you back β€οΈ
gets logged.
console.log(String.raw`Hello\nworld`);
- A:
Hello world!
- B:
Hello
Β Β Βworld
- C:
Hello\nworld
- D:
Hello\n
Β Β Βworld
Answer
String.raw
returns a string where the escapes (\n
, \v
, \t
etc.) are ignored! Backslashes can be an issue since you could end up with something like:
const path = `C:\Documents\Projects\table.html`
Which would result in:
"C:DocumentsProjects able.html"
With String.raw
, it would simply ignore the escape and print:
C:\Documents\Projects\table.html
In this case, the string is Hello\nworld
, which gets logged.
async function getData() {
return await Promise.resolve('I made it!');
}
const data = getData();
console.log(data);
- A:
"I made it!"
- B:
Promise {<resolved>: "I made it!"}
- C:
Promise {<pending>}
- D:
undefined
Answer
An async function always returns a promise. The await
still has to wait for the promise to resolve: a pending promise gets returned when we call getData()
in order to set data
equal to it.
If we wanted to get access to the resolved value "I made it"
, we could have used the .then()
method on data
:
data.then(res => console.log(res))
This would've logged "I made it!"
function addToList(item, list) {
return list.push(item);
}
const result = addToList('apple', ['banana']);
console.log(result);
- A:
['apple', 'banana']
- B:
2
- C:
true
- D:
undefined
Answer
The .push()
method returns the length of the new array! Previously, the array contained one element (the string "banana"
) and had a length of 1
. After adding the string "apple"
to the array, the array contains two elements, and has a length of 2
. This gets returned from the addToList
function.
The push
method modifies the original array. If you wanted to return the array from the function rather than the length of the array, you should have returned list
after pushing item
to it.
const box = { x: 10, y: 20 };
Object.freeze(box);
const shape = box;
shape.x = 100;
console.log(shape);
- A:
{ x: 100, y: 20 }
- B:
{ x: 10, y: 20 }
- C:
{ x: 100 }
- D:
ReferenceError
Answer
Object.freeze
makes it impossible to add, remove, or modify properties of an object (unless the property's value is another object).
When we create the variable shape
and set it equal to the frozen object box
, shape
also refers to a frozen object. You can check whether an object is frozen by using Object.isFrozen
. In this case, Object.isFrozen(shape)
returns true, since the variable shape
has a reference to a frozen object.
Since shape
is frozen, and since the value of x
is not an object, we cannot modify the property x
. x
is still equal to 10
, and { x: 10, y: 20 }
gets logged.
const { name: myName } = { name: 'Lydia' };
console.log(name);
- A:
"Lydia"
- B:
"myName"
- C:
undefined
- D:
ReferenceError
Answer
When we unpack the property name
from the object on the right-hand side, we assign its value "Lydia"
to a variable with the name myName
.
With { name: myName }
, we tell JavaScript that we want to create a new variable called myName
with the value of the name
property on the right-hand side.
Since we try to log name
, a variable that is not defined, a ReferenceError gets thrown.
function sum(a, b) {
return a + b;
}
- A: Yes
- B: No
Answer
A pure function is a function that always returns the same result, if the same arguments are passed.
The sum
function always returns the same result. If we pass 1
and 2
, it will always return 3
without side effects. If we pass 5
and 10
, it will always return 15
, and so on. This is the definition of a pure function.
const add = () => {
const cache = {};
return num => {
if (num in cache) {
return `From cache! ${cache[num]}`;
} else {
const result = num + 10;
cache[num] = result;
return `Calculated! ${result}`;
}
};
};
const addFunction = add();
console.log(addFunction(10));
console.log(addFunction(10));
console.log(addFunction(5 * 2));
- A:
Calculated! 20
Calculated! 20
Calculated! 20
- B:
Calculated! 20
From cache! 20
Calculated! 20
- C:
Calculated! 20
From cache! 20
From cache! 20
- D:
Calculated! 20
From cache! 20
Error
Answer
The add
function is a memoized function. With memoization, we can cache the results of a function in order to speed up its execution. In this case, we create a cache
object that stores the previously returned values.
If we call the addFunction
function again with the same argument, it first checks whether it has already gotten that value in its cache. If that's the case, the caches value will be returned, which saves on execution time. Else, if it's not cached, it will calculate the value and store it afterwards.
We call the addFunction
function three times with the same value: on the first invocation, the value of the function when num
is equal to 10
isn't cached yet. The condition of the if-statement num in cache
returns false
, and the else block gets executed: Calculated! 20
gets logged, and the value of the result gets added to the cache object. cache
now looks like { 10: 20 }
.
The second time, the cache
object contains the value that gets returned for 10
. The condition of the if-statement num in cache
returns true
, and 'From cache! 20'
gets logged.
The third time, we pass 5 * 2
to the function which gets evaluated to 10
. The cache
object contains the value that gets returned for 10
. The condition of the if-statement num in cache
returns true
, and 'From cache! 20'
gets logged.
const myLifeSummedUp = ['β', 'π»', 'π·', 'π«'];
for (let item in myLifeSummedUp) {
console.log(item);
}
for (let item of myLifeSummedUp) {
console.log(item);
}
- A:
0
1
2
3
and"β"
"π»"
"π·"
"π«"
- B:
"β"
"π»"
"π·"
"π«"
and"β"
"π»"
"π·"
"π«"
- C:
"β"
"π»"
"π·"
"π«"
and0
1
2
3
- D:
0
1
2
3
and{0: "β", 1: "π»", 2: "π·", 3: "π«"}
Answer
With a for-in loop, we can iterate over enumerable properties. In an array, the enumerable properties are the "keys" of array elements, which are actually their indexes. You could see an array as:
{0: "β", 1: "π»", 2: "π·", 3: "π«"}
Where the keys are the enumerable properties. 0
1
2
3
get logged.
With a for-of loop, we can iterate over iterables. An array is an iterable. When we iterate over the array, the variable "item" is equal to the element it's currently iterating over, "β"
"π»"
"π·"
"π«"
get logged.
const list = [1 + 2, 1 * 2, 1 / 2];
console.log(list);
- A:
["1 + 2", "1 * 2", "1 / 2"]
- B:
["12", 2, 0.5]
- C:
[3, 2, 0.5]
- D:
[1, 1, 1]
Answer
Array elements can hold any value. Numbers, strings, objects, other arrays, null, boolean values, undefined, and other expressions such as dates, functions, and calculations.
The element will be equal to the returned value. 1 + 2
returns 3
, 1 * 2
returns 2
, and 1 / 2
returns 0.5
.
function sayHi(name) {
return `Hi there, ${name}`;
}
console.log(sayHi());
- A:
Hi there,
- B:
Hi there, undefined
- C:
Hi there, null
- D:
ReferenceError
Answer
By default, arguments have the value of undefined
, unless a value has been passed to the function. In this case, we didn't pass a value for the name
argument. name
is equal to undefined
which gets logged.
In ES6, we can overwrite this default undefined
value with default parameters. For example:
function sayHi(name = "Lydia") { ... }
In this case, if we didn't pass a value or if we passed undefined
, name
would always be equal to the string Lydia
var status = 'π';
setTimeout(() => {
const status = 'π';
const data = {
status: 'π₯',
getStatus() {
return this.status;
},
};
console.log(data.getStatus());
console.log(data.getStatus.call(this));
}, 0);
- A:
"π₯"
and"π"
- B:
"π₯"
and"π"
- C:
"π"
and"π"
- D:
"π"
and"π"
Answer
The value of the this
keyword is dependent on where you use it. In a method, like the getStatus
method, the this
keyword refers to the object that the method belongs to. The method belongs to the data
object, so this
refers to the data
object. When we log this.status
, the status
property on the data
object gets logged, which is "π₯"
.
With the call
method, we can change the object to which the this
keyword refers. In functions, the this
keyword refers to the the object that the function belongs to. We declared the setTimeout
function on the global object, so within the setTimeout
function, the this
keyword refers to the global object. On the global object, there is a variable called status with the value of "π"
. When logging this.status
, "π"
gets logged.
const person = {
name: 'Lydia',
age: 21,
};
let city = person.city;
city = 'Amsterdam';
console.log(person);
- A:
{ name: "Lydia", age: 21 }
- B:
{ name: "Lydia", age: 21, city: "Amsterdam" }
- C:
{ name: "Lydia", age: 21, city: undefined }
- D:
"Amsterdam"
Answer
We set the variable city
equal to the value of the property called city
on the person
object. There is no property on this object called city
, so the variable city
has the value of undefined
.
Note that we are not referencing the person
object itself! We simply set the variable city
equal to the current value of the city
property on the person
object.
Then, we set city
equal to the string "Amsterdam"
. This doesn't change the person object: there is no reference to that object.
When logging the person
object, the unmodified object gets returned.
function checkAge(age) {
if (age < 18) {
const message = "Sorry, you're too young.";
} else {
const message = "Yay! You're old enough!";
}
return message;
}
console.log(checkAge(21));
- A:
"Sorry, you're too young."
- B:
"Yay! You're old enough!"
- C:
ReferenceError
- D:
undefined
Answer
Variables with the const
and let
keyword are block-scoped. A block is anything between curly brackets ({ }
). In this case, the curly brackets of the if/else statements. You cannot reference a variable outside of the block it's declared in, a ReferenceError gets thrown.
fetch('https://www.website.com/api/user/1')
.then(res => res.json())
.then(res => console.log(res));
- A: The result of the
fetch
method. - B: The result of the second invocation of the
fetch
method. - C: The result of the callback in the previous
.then()
. - D: It would always be undefined.
Answer
The value of res
in the second .then
is equal to the returned value of the previous .then
. You can keep chaining .then
s like this, where the value is passed to the next handler.
86. Which option is a way to set hasName
equal to true
, provided you cannot pass true
as an argument?
function getName(name) {
const hasName = //
}
- A:
!!name
- B:
name
- C:
new Boolean(name)
- D:
name.length
Answer
With !!name
, we determine whether the value of name
is truthy or falsy. If name is truthy, which we want to test for, !name
returns false
. !false
(which is what !!name
practically is) returns true
.
By setting hasName
equal to name
, you set hasName
equal to whatever value you passed to the getName
function, not the boolean value true
.
new Boolean(true)
returns an object wrapper, not the boolean value itself.
name.length
returns the length of the passed argument, not whether it's true
.
console.log('I want pizza'[0]);
- A:
"""
- B:
"I"
- C:
SyntaxError
- D:
undefined
Answer
In order to get an character on a specific index in a string, you can use bracket notation. The first character in the string has index 0, and so on. In this case we want to get the element which index is 0, the character "I'
, which gets logged.
Note that this method is not supported in IE7 and below. In that case, use .charAt()
function sum(num1, num2 = num1) {
console.log(num1 + num2);
}
sum(10);
- A:
NaN
- B:
20
- C:
ReferenceError
- D:
undefined
Answer
You can set a default parameter's value equal to another parameter of the function, as long as they've been defined before the default parameter. We pass the value 10
to the sum
function. If the sum
function only receives 1 argument, it means that the value for num2
is not passed, and the value of num1
is equal to the passed value 10
in this case. The default value of num2
is the value of num1
, which is 10
. num1 + num2
returns 20
.
If you're trying to set a default parameter's value equal to a parameter which is defined after (to the right), the parameter's value hasn't been initialized yet, which will throw an error.
// module.js
export default () => 'Hello world';
export const name = 'Lydia';
// index.js
import * as data from './module';
console.log(data);
- A:
{ default: function default(), name: "Lydia" }
- B:
{ default: function default() }
- C:
{ default: "Hello world", name: "Lydia" }
- D: Global object of
module.js
Answer
With the import * as name
syntax, we import all exports from the module.js
file into the index.js
file as a new object called data
is created. In the module.js
file, there are two exports: the default export, and a named export. The default export is a function which returns the string "Hello World"
, and the named export is a variable called name
which has the value of the string "Lydia"
.
The data
object has a default
property for the default export, other properties have the names of the named exports and their corresponding values.
class Person {
constructor(name) {
this.name = name;
}
}
const member = new Person('John');
console.log(typeof member);
- A:
"class"
- B:
"function"
- C:
"object"
- D:
"string"
Answer
Classes are syntactical sugar for function constructors. The equivalent of the Person
class as a function constructor would be:
function Person() {
this.name = name;
}
Calling a function constructor with new
results in the creation of an instance of Person
, typeof
keyword returns "object"
for an instance. typeof member
returns "object"
.
let newList = [1, 2, 3].push(4);
console.log(newList.push(5));
- A:
[1, 2, 3, 4, 5]
- B:
[1, 2, 3, 5]
- C:
[1, 2, 3, 4]
- D:
Error
Answer
The .push
method returns the new length of the array, not the array itself! By setting newList
equal to [1, 2, 3].push(4)
, we set newList
equal to the new length of the array: 4
.
Then, we try to use the .push
method on newList
. Since newList
is the numerical value 4
, we cannot use the .push
method: a TypeError is thrown.
function giveLydiaPizza() {
return 'Here is pizza!';
}
const giveLydiaChocolate = () =>
"Here's chocolate... now go hit the gym already.";
console.log(giveLydiaPizza.prototype);
console.log(giveLydiaChocolate.prototype);
- A:
{ constructor: ...}
{ constructor: ...}
- B:
{}
{ constructor: ...}
- C:
{ constructor: ...}
{}
- D:
{ constructor: ...}
undefined
Answer
Regular functions, such as the giveLydiaPizza
function, have a prototype
property, which is an object (prototype object) with a constructor
property. Arrow functions however, such as the giveLydiaChocolate
function, do not have this prototype
property. undefined
gets returned when trying to access the prototype
property using giveLydiaChocolate.prototype
.
const person = {
name: 'Lydia',
age: 21,
};
for (const [x, y] of Object.entries(person)) {
console.log(x, y);
}
- A:
name
Lydia
andage
21
- B:
["name", "Lydia"]
and["age", 21]
- C:
["name", "age"]
andundefined
- D:
Error
Answer
Object.entries(person)
returns an array of nested arrays, containing the keys and objects:
[ [ 'name', 'Lydia' ], [ 'age', 21 ] ]
Using the for-of
loop, we can iterate over each element in the array, the subarrays in this case. We can destructure the subarrays instantly in the for-of loop, using const [x, y]
. x
is equal to the first element in the subarray, y
is equal to the second element in the subarray.
The first subarray is [ "name", "Lydia" ]
, with x
equal to "name"
, and y
equal to "Lydia"
, which get logged.
The second subarray is [ "age", 21 ]
, with x
equal to "age"
, and y
equal to 21
, which get logged.
function getItems(fruitList, ...args, favoriteFruit) {
return [...fruitList, ...args, favoriteFruit]
}
getItems(["banana", "apple"], "pear", "orange")
- A:
["banana", "apple", "pear", "orange"]
- B:
[["banana", "apple"], "pear", "orange"]
- C:
["banana", "apple", ["pear"], "orange"]
- D:
SyntaxError
Answer
...args
is a rest parameter. The rest parameter's value is an array containing all remaining arguments, and can only be the last parameter! In this example, the rest parameter was the second parameter. This is not possible, and will throw a syntax error.
function getItems(fruitList, favoriteFruit, ...args) {
return [...fruitList, ...args, favoriteFruit];
}
getItems(['banana', 'apple'], 'pear', 'orange');
The above example works. This returns the array [ 'banana', 'apple', 'orange', 'pear' ]
function nums(a, b) {
if (a > b) console.log('a is bigger');
else console.log('b is bigger');
return;
a + b;
}
console.log(nums(4, 2));
console.log(nums(1, 2));
- A:
a is bigger
,6
andb is bigger
,3
- B:
a is bigger
,undefined
andb is bigger
,undefined
- C:
undefined
andundefined
- D:
SyntaxError
Answer
In JavaScript, we don't have to write the semicolon (;
) explicitly, however the JavaScript engine still adds them after statements. This is called Automatic Semicolon Insertion. A statement can for example be variables, or keywords like throw
, return
, break
, etc.
Here, we wrote a return
statement, and another value a + b
on a new line. However, since it's a new line, the engine doesn't know that it's actually the value that we wanted to return. Instead, it automatically added a semicolon after return
. You could see this as:
return;
a + b;
This means that a + b
is never reached, since a function stops running after the return
keyword. If no value gets returned, like here, the function returns undefined
. Note that there is no automatic insertion after if/else
statements!
class Person {
constructor() {
this.name = 'Lydia';
}
}
Person = class AnotherPerson {
constructor() {
this.name = 'Sarah';
}
};
const member = new Person();
console.log(member.name);
- A:
"Lydia"
- B:
"Sarah"
- C:
Error: cannot redeclare Person
- D:
SyntaxError
Answer
We can set classes equal to other classes/function constructors. In this case, we set Person
equal to AnotherPerson
. The name on this constructor is Sarah
, so the name property on the new Person
instance member
is "Sarah"
.
const info = {
[Symbol('a')]: 'b',
};
console.log(info);
console.log(Object.keys(info));
- A:
{Symbol('a'): 'b'}
and["{Symbol('a')"]
- B:
{}
and[]
- C:
{ a: "b" }
and["a"]
- D:
{Symbol('a'): 'b'}
and[]
Answer
A Symbol is not enumerable. The Object.keys method returns all enumerable key properties on an object. The Symbol won't be visible, and an empty array is returned. When logging the entire object, all properties will be visible, even non-enumerable ones.
This is one of the many qualities of a symbol: besides representing an entirely unique value (which prevents accidental name collision on objects, for example when working with 2 libraries that want to add properties to the same object), you can also "hide" properties on objects this way (although not entirely. You can still access symbols using the Object.getOwnPropertySymbols()
method).
const getList = ([x, ...y]) => [x, y]
const getUser = user => { name: user.name, age: user.age }
const list = [1, 2, 3, 4]
const user = { name: "Lydia", age: 21 }
console.log(getList(list))
console.log(getUser(user))
- A:
[1, [2, 3, 4]]
andundefined
- B:
[1, [2, 3, 4]]
and{ name: "Lydia", age: 21 }
- C:
[1, 2, 3, 4]
and{ name: "Lydia", age: 21 }
- D:
Error
and{ name: "Lydia", age: 21 }
Answer
The getList
function receives an array as its argument. Between the parentheses of the getList
function, we destructure this array right away. You could see this as:
[x, ...y] = [1, 2, 3, 4]
With the rest parameter ...y
, we put all "remaining" arguments in an array. The remaining arguments are 2
, 3
and 4
in this case. The value of y
is an array, containing all the rest parameters. The value of x
is equal to 1
in this case, so when we log [x, y]
, [1, [2, 3, 4]]
gets logged.
The getUser
function receives an object. With arrow functions, we don't have to write curly brackets if we just return one value. However, if you want to return an object from an arrow function, you have to write it between parentheses, otherwise no value gets returned! The following function would have returned an object:
const getUser = user => ({ name: user.name, age: user.age })
Since no value gets returned in this case, the function returns undefined
.
const name = 'Lydia';
console.log(name());
- A:
SyntaxError
- B:
ReferenceError
- C:
TypeError
- D:
undefined
Answer
The variable name
holds the value of a string, which is not a function, thus cannot invoke.
TypeErrors get thrown when a value is not of the expected type. JavaScript expected name
to be a function since we're trying to invoke it. It was a string however, so a TypeError gets thrown: name is not a function!
SyntaxErrors get thrown when you've written something that isn't valid JavaScript, for example when you've written the word return
as retrun
.
ReferenceErrors get thrown when JavaScript isn't able to find a reference to a value that you're trying to access.
// πβ¨ This is my 100th question! β¨π
const output = `${[] && 'Im'}possible!
You should${'' && `n't`} see a therapist after so much JavaScript lol`;
- A:
possible! You should see a therapist after so much JavaScript lol
- B:
Impossible! You should see a therapist after so much JavaScript lol
- C:
possible! You shouldn't see a therapist after so much JavaScript lol
- D:
Impossible! You shouldn't see a therapist after so much JavaScript lol
Answer
[]
is a truthy value. With the &&
operator, the right-hand value will be returned if the left-hand value is a truthy value. In this case, the left-hand value []
is a truthy value, so "Im'
gets returned.
""
is a falsy value. If the left-hand value is falsy, nothing gets returned. n't
doesn't get returned.
const one = false || {} || null;
const two = null || false || '';
const three = [] || 0 || true;
console.log(one, two, three);
- A:
false
null
[]
- B:
null
""
true
- C:
{}
""
[]
- D:
null
null
true
Answer
With the ||
operator, we can return the first truthy operand. If all values are falsy, the last operand gets returned.
(false || {} || null)
: the empty object {}
is a truthy value. This is the first (and only) truthy value, which gets returned. one
is equal to {}
.
(null || false || "")
: all operands are falsy values. This means that the last operand, ""
gets returned. two
is equal to ""
.
([] || 0 || "")
: the empty array[]
is a truthy value. This is the first truthy value, which gets returned. three
is equal to []
.
const myPromise = () => Promise.resolve('I have resolved!');
function firstFunction() {
myPromise().then(res => console.log(res));
console.log('second');
}
async function secondFunction() {
console.log(await myPromise());
console.log('second');
}
firstFunction();
secondFunction();
- A:
I have resolved!
,second
andI have resolved!
,second
- B:
second
,I have resolved!
andsecond
,I have resolved!
- C:
I have resolved!
,second
andsecond
,I have resolved!
- D:
second
,I have resolved!
andI have resolved!
,second
Answer
With a promise, we basically say I want to execute this function, but I'll put it aside for now while it's running since this might take a while. Only when a certain value is resolved (or rejected), and when the call stack is empty, I want to use this value.
We can get this value with both .then
and the await
keyword in an async
function. Although we can get a promise's value with both .then
and await
, they work a bit differently.
In the firstFunction
, we (sort of) put the myPromise function aside while it was running, but continued running the other code, which is console.log('second')
in this case. Then, the function resolved with the string I have resolved
, which then got logged after it saw that the callstack was empty.
With the await keyword in secondFunction
, we literally pause the execution of an async function until the value has been resolved before moving to the next line.
This means that it waited for the myPromise
to resolve with the value I have resolved
, and only once that happened, we moved to the next line: second
got logged.
const set = new Set();
set.add(1);
set.add('Lydia');
set.add({ name: 'Lydia' });
for (let item of set) {
console.log(item + 2);
}
- A:
3
,NaN
,NaN
- B:
3
,7
,NaN
- C:
3
,Lydia2
,[object Object]2
- D:
"12"
,Lydia2
,[object Object]2
Answer
The +
operator is not only used for adding numerical values, but we can also use it to concatenate strings. Whenever the JavaScript engine sees that one or more values are not a number, it coerces the number into a string.
The first one is 1
, which is a numerical value. 1 + 2
returns the number 3.
However, the second one is a string "Lydia"
. "Lydia"
is a string and 2
is a number: 2
gets coerced into a string. "Lydia"
and "2"
get concatenated, which results in the string "Lydia2"
.
{ name: "Lydia" }
is an object. Neither a number nor an object is a string, so it stringifies both. Whenever we stringify a regular object, it becomes "[object Object]"
. "[object Object]"
concatenated with "2"
becomes "[object Object]2"
.
Promise.resolve(5);
- A:
5
- B:
Promise {<pending>: 5}
- C:
Promise {<resolved>: 5}
- D:
Error
Answer
We can pass any type of value we want to Promise.resolve
, either a promise or a non-promise. The method itself returns a promise with the resolved value. If you pass a regular function, it'll be a resolved promise with a regular value. If you pass a promise, it'll be a resolved promise with the resolved value of that passed promise.
In this case, we just passed the numerical value 5
. It returns a resolved promise with the value 5
.
function compareMembers(person1, person2 = person) {
if (person1 !== person2) {
console.log('Not the same!');
} else {
console.log('They are the same!');
}
}
const person = { name: 'Lydia' };
compareMembers(person);
- A:
Not the same!
- B:
They are the same!
- C:
ReferenceError
- D:
SyntaxError
Answer
Objects are passed by reference. When we check objects for strict equality (===
), we're comparing their references.
We set the default value for person2
equal to the person
object, and passed the person
object as the value for person1
.
This means that both values have a reference to the same spot in memory, thus they are equal.
The code block in the else
statement gets run, and They are the same!
gets logged.
const colorConfig = {
red: true,
blue: false,
green: true,
black: true,
yellow: false,
};
const colors = ['pink', 'red', 'blue'];
console.log(colorConfig.colors[1]);
- A:
true
- B:
false
- C:
undefined
- D:
TypeError
Answer
In JavaScript, we have two ways to access properties on an object: bracket notation, or dot notation. In this example, we use dot notation (colorConfig.colors
) instead of bracket notation (colorConfig["colors"]
).
With dot notation, JavaScript tries to find the property on the object with that exact name. In this example, JavaScript tries to find a property called colors
on the colorConfig
object. There is no property called colors
, so this returns undefined
. Then, we try to access the value of the first element by using [1]
. We cannot do this on a value that's undefined
, so it throws a TypeError
: Cannot read property '1' of undefined
.
JavaScript interprets (or unboxes) statements. When we use bracket notation, it sees the first opening bracket [
and keeps going until it finds the closing bracket ]
. Only then, it will evaluate the statement. If we would've used colorConfig[colors[1]]
, it would have returned the value of the red
property on the colorConfig
object.
console.log('β€οΈ' === 'β€οΈ');
- A:
true
- B:
false
Answer
Under the hood, emojis are unicodes. The unicodes for the heart emoji is "U+2764 U+FE0F"
. These are always the same for the same emojis, so we're comparing two equal strings to each other, which returns true.
const emojis = ['β¨', 'π₯', 'π'];
emojis.map(x => x + 'β¨');
emojis.filter(x => x !== 'π₯');
emojis.find(x => x !== 'π₯');
emojis.reduce((acc, cur) => acc + 'β¨');
emojis.slice(1, 2, 'β¨');
emojis.splice(1, 2, 'β¨');
- A:
All of them
- B:
map
reduce
slice
splice
- C:
map
slice
splice
- D:
splice
Answer
With splice
method, we modify the original array by deleting, replacing or adding elements. In this case, we removed 2 items from index 1 (we removed 'π₯'
and 'π'
) and added the β¨ emoji instead.
map
, filter
and slice
return a new array, find
returns an element, and reduce
returns a reduced value.
const food = ['π', 'π«', 'π₯', 'π'];
const info = { favoriteFood: food[0] };
info.favoriteFood = 'π';
console.log(food);
- A:
['π', 'π«', 'π₯', 'π']
- B:
['π', 'π«', 'π₯', 'π']
- C:
['π', 'π', 'π«', 'π₯', 'π']
- D:
ReferenceError
Answer
We set the value of the favoriteFood
property on the info
object equal to the string with the pizza emoji, 'π'
. A string is a primitive data type. In JavaScript, primitive data types act by reference
In JavaScript, primitive data types (everything that's not an object) interact by value. In this case, we set the value of the favoriteFood
property on the info
object equal to the value of the first element in the food
array, the string with the pizza emoji in this case ('π'
). A string is a primitive data type, and interact by value (see my blogpost if you're interested in learning more)
Then, we change the value of the favoriteFood
property on the info
object. The food
array hasn't changed, since the value of favoriteFood
was merely a copy of the value of the first element in the array, and doesn't have a reference to the same spot in memory as the element on food[0]
. When we log food, it's still the original array, ['π', 'π«', 'π₯', 'π']
.
JSON.parse();
- A: Parses JSON to a JavaScript value
- B: Parses a JavaScript object to JSON
- C: Parses any JavaScript value to JSON
- D: Parses JSON to a JavaScript object only
Answer
With the JSON.parse()
method, we can parse JSON string to a JavaScript value.
// Stringifying a number into valid JSON, then parsing the JSON string to a JavaScript value:
const jsonNumber = JSON.stringify(4); // '4'
JSON.parse(jsonNumber); // 4
// Stringifying an array value into valid JSON, then parsing the JSON string to a JavaScript value:
const jsonArray = JSON.stringify([1, 2, 3]); // '[1, 2, 3]'
JSON.parse(jsonArray); // [1, 2, 3]
// Stringifying an object into valid JSON, then parsing the JSON string to a JavaScript value:
const jsonArray = JSON.stringify({ name: 'Lydia' }); // '{"name":"Lydia"}'
JSON.parse(jsonArray); // { name: 'Lydia' }
let name = 'Lydia';
function getName() {
console.log(name);
let name = 'Sarah';
}
getName();
- A: Lydia
- B: Sarah
- C:
undefined
- D:
ReferenceError
Answer
Each function has its own execution context (or scope). The getName
function first looks within its own context (scope) to see if it contains the variable name
we're trying to access. In this case, the getName
function contains its own name
variable: we declare the variable name
with the let
keyword, and with the value of 'Sarah'
.
Variables with the let
keyword (and const
) are hoisted, but unlike var
, don't get initialized. They are not accessible before the line we declare (initialize) them. This is called the "temporal dead zone". When we try to access the variables before they are declared, JavaScript throws a ReferenceError
.
If we wouldn't have declared the name
variable within the getName
function, the javascript engine would've looked down the scope chain. The outer scope has a variable called name
with the value of Lydia
. In that case, it would've logged Lydia
.
let name = 'Lydia';
function getName() {
console.log(name);
}
getName(); // Lydia
function* generatorOne() {
yield ['a', 'b', 'c'];
}
function* generatorTwo() {
yield* ['a', 'b', 'c'];
}
const one = generatorOne();
const two = generatorTwo();
console.log(one.next().value);
console.log(two.next().value);
- A:
a
anda
- B:
a
andundefined
- C:
['a', 'b', 'c']
anda
- D:
a
and['a', 'b', 'c']
Answer
With the yield
keyword, we yield
values in a generator function. With the yield*
keyword, we can yield values from another generator function, or iterable object (for example an array).
In generatorOne
, we yield the entire array ['a', 'b', 'c']
using the yield
keyword. The value of value
property on the object returned by the next
method on one
(one.next().value
) is equal to the entire array ['a', 'b', 'c']
.
console.log(one.next().value); // ['a', 'b', 'c']
console.log(one.next().value); // undefined
In generatorTwo
, we use the yield*
keyword. This means that the first yielded value of two
, is equal to the first yielded value in the iterator. The iterator is the array ['a', 'b', 'c']
. The first yielded value is a
, so the first time we call two.next().value
, a
is returned.
console.log(two.next().value); // 'a'
console.log(two.next().value); // 'b'
console.log(two.next().value); // 'c'
console.log(two.next().value); // undefined
console.log(`${(x => x)('I love')} to program`);
- A:
I love to program
- B:
undefined to program
- C:
${(x => x)('I love') to program
- D:
TypeError
Answer
Expressions within template literals are evaluated first. This means that the string will contain the returned value of the expression, the immediately invoked function (x => x)('I love')
in this case. We pass the value 'I love'
as an argument to the x => x
arrow function. x
is equal to 'I love'
, which gets returned. This results in I love to program
.
let config = {
alert: setInterval(() => {
console.log('Alert!');
}, 1000),
};
config = null;
- A: The
setInterval
callback won't be invoked - B: The
setInterval
callback gets invoked once - C: The
setInterval
callback will still be called every second - D: We never invoked
config.alert()
, config isnull
Answer
Normally when we set objects equal to null
, those objects get garbage collected as there is no reference anymore to that object. However, since the callback function within setInterval
is an arrow function (thus bound to the config
object), the callback function still holds a reference to the config
object. As long as there is a reference, the object won't get garbage collected. Since it's not garbage collected, the setInterval
callback function will still get invoked every 1000ms (1s).
const myMap = new Map();
const myFunc = () => 'greeting';
myMap.set(myFunc, 'Hello world!');
//1
myMap.get('greeting');
//2
myMap.get(myFunc);
//3
myMap.get(() => 'greeting');
- A: 1
- B: 2
- C: 2 and 3
- D: All of them
Answer
When adding a key/value pair using the set
method, the key will be the value of the first argument passed to the set
function, and the value will be the second argument passed to the set
function. The key is the function () => 'greeting'
in this case, and the value 'Hello world'
. myMap
is now { () => 'greeting' => 'Hello world!' }
.
1 is wrong, since the key is not 'greeting'
but () => 'greeting'
.
3 is wrong, since we're creating a new function by passing it as a parameter to the get
method. Object interact by reference. Functions are objects, which is why two functions are never strictly equal, even if they are identical: they have a reference to a different spot in memory.
const person = {
name: 'Lydia',
age: 21,
};
const changeAge = (x = { ...person }) => (x.age += 1);
const changeAgeAndName = (x = { ...person }) => {
x.age += 1;
x.name = 'Sarah';
};
changeAge(person);
changeAgeAndName();
console.log(person);
- A:
{name: "Sarah", age: 22}
- B:
{name: "Sarah", age: 23}
- C:
{name: "Lydia", age: 22}
- D:
{name: "Lydia", age: 23}
Answer
Both the changeAge
and changeAgeAndName
functions have a default parameter, namely a newly created object { ...person }
. This object has copies of all the key/values in the person
object.
First, we invoke the changeAge
function and pass the person
object as its argument. This function increases the value of the age
property by 1. person
is now { name: "Lydia", age: 22 }
.
Then, we invoke the changeAgeAndName
function, however we don't pass a parameter. Instead, the value of x
is equal to a new object: { ...person }
. Since it's a new object, it doesn't affect the values of the properties on the person
object. person
is still equal to { name: "Lydia", age: 22 }
.
function sumValues(x, y, z) {
return x + y + z;
}
- A:
sumValues([...1, 2, 3])
- B:
sumValues([...[1, 2, 3]])
- C:
sumValues(...[1, 2, 3])
- D:
sumValues([1, 2, 3])
Answer
With the spread operator ...
, we can spread iterables to individual elements. The sumValues
function receives three arguments: x
, y
and z
. ...[1, 2, 3]
will result in 1, 2, 3
, which we pass to the sumValues
function.
let num = 1;
const list = ['π₯³', 'π€ ', 'π₯°', 'π€ͺ'];
console.log(list[(num += 1)]);
- A:
π€
- B:
π₯°
- C:
SyntaxError
- D:
ReferenceError
Answer
With the +=
operand, we're incrementing the value of num
by 1
. num
had the initial value 1
, so 1 + 1
is 2
. The item on the second index in the list
array is π₯°, console.log(list[2])
prints π₯°.
const person = {
firstName: 'Lydia',
lastName: 'Hallie',
pet: {
name: 'Mara',
breed: 'Dutch Tulip Hound',
},
getFullName() {
return `${this.firstName} ${this.lastName}`;
},
};
console.log(person.pet?.name);
console.log(person.pet?.family?.name);
console.log(person.getFullName?.());
console.log(member.getLastName?.());
- A:
undefined
undefined
undefined
undefined
- B:
Mara
undefined
Lydia Hallie
undefined
- C:
Mara
null
Lydia Hallie
null
- D:
null
ReferenceError
null
ReferenceError
Answer
With the optional chaining operator ?.
, we no longer have to explicitly check whether the deeper nested values are valid or not. If we're trying to access a property on an undefined
or null
value (nullish), the expression short-circuits and returns undefined
.
person.pet?.name
: person
has a property named pet
: person.pet
is not nullish. It has a property called name
, and returns Mara
.
person.pet?.family?.name
: person
has a property named pet
: person.pet
is not nullish. pet
does not have a property called family
, person.pet.family
is nullish. The expression returns undefined
.
person.getFullName?.()
: person
has a property named getFullName
: person.getFullName()
is not nullish and can get invoked, which returns Lydia Hallie
.
member.getLastName?.()
: member
is not defined: member.getLastName()
is nullish. The expression returns undefined
.
const groceries = ['banana', 'apple', 'peanuts'];
if (groceries.indexOf('banana')) {
console.log('We have to buy bananas!');
} else {
console.log(`We don't have to buy bananas!`);
}
- A: We have to buy bananas!
- B: We don't have to buy bananas
- C:
undefined
- D:
1
Answer
We passed the condition groceries.indexOf("banana")
to the if-statement. groceries.indexOf("banana")
returns 0
, which is a falsy value. Since the condition in the if-statement is falsy, the code in the else
block runs, and We don't have to buy bananas!
gets logged.
const config = {
languages: [],
set language(lang) {
return this.languages.push(lang);
},
};
console.log(config.language);
- A:
function language(lang) { this.languages.push(lang }
- B:
0
- C:
[]
- D:
undefined
Answer
The language
method is a setter
. Setters don't hold an actual value, their purpose is to modify properties. When calling a setter
method, undefined
gets returned.
const name = 'Lydia Hallie';
console.log(!typeof name === 'object');
console.log(!typeof name === 'string');
- A:
false
true
- B:
true
false
- C:
false
false
- D:
true
true
Answer
typeof name
returns "string"
. The string "string"
is a truthy value, so !typeof name
returns the boolean value false
. false === "object"
and false === "string"
both returnfalse
.
(If we wanted to check whether the type was (un)equal to a certain type, we should've written !==
instead of !typeof
)
const add = x => y => z => {
console.log(x, y, z);
return x + y + z;
};
add(4)(5)(6);
- A:
4
5
6
- B:
6
5
4
- C:
4
function
function
- D:
undefined
undefined
6
Answer
The add
function returns an arrow function, which returns an arrow function, which returns an arrow function (still with me?). The first function receives an argument x
with the value of 4
. We invoke the second function, which receives an argument y
with the value 5
. Then we invoke the third function, which receives an argument z
with the value 6
. When we're trying to access the value x
, y
and z
within the last arrow function, the JS engine goes up the scope chain in order to find the values for x
and y
accordingly. This returns 4
5
6
.
async function* range(start, end) {
for (let i = start; i <= end; i++) {
yield Promise.resolve(i);
}
}
(async () => {
const gen = range(1, 3);
for await (const item of gen) {
console.log(item);
}
})();
- A:
Promise {1}
Promise {2}
Promise {3}
- B:
Promise {<pending>}
Promise {<pending>}
Promise {<pending>}
- C:
1
2
3
- D:
undefined
undefined
undefined
Answer
The generator function range
returns an async object with promises for each item in the range we pass: Promise{1}
, Promise{2}
, Promise{3}
. We set the variable gen
equal to the async object, after which we loop over it using a for await ... of
loop. We set the variable item
equal to the returned Promise values: first Promise{1}
, then Promise{2}
, then Promise{3}
. Since we're awaiting the value of item
, the resolved promsie, the resolved values of the promises get returned: 1
, 2
, then 3
.
const myFunc = ({ x, y, z }) => {
console.log(x, y, z);
};
myFunc(1, 2, 3);
- A:
1
2
3
- B:
{1: 1}
{2: 2}
{3: 3}
- C:
{ 1: undefined }
undefined
undefined
- D:
undefined
undefined
undefined
Answer
myFunc
expects an object with properties x
, y
and z
as its argument. Since we're only passing three separate numeric values (1, 2, 3) instead of one object with properties x
, y
and z
({x: 1, y: 2, z: 3}), x
, y
and z
have their default value of undefined
.
function getFine(speed, amount) {
const formattedSpeed = new Intl.NumberFormat({
'en-US',
{ style: 'unit', unit: 'mile-per-hour' }
}).format(speed)
const formattedAmount = new Intl.NumberFormat({
'en-US',
{ style: 'currency', currency: 'USD' }
}).format(amount)
return `The driver drove ${formattedSpeed} and has to pay ${formattedAmount}`
}
console.log(getFine(130, 300))
- A: The driver drove 130 and has to pay 300
- B: The driver drove 130 mph and has to pay $300.00
- C: The driver drove undefined and has to pay undefined
- D: The driver drove 130.00 and has to pay 300.00
Answer
With the Intl.NumberFormat
method, we can format numeric values to any locale. We format the numeric value 130
to the en-US
locale as a unit
in mile-per-hour
, which results in 130 mph
. The numeric value 300
to the en-US
locale as a currentcy
in USD
results in $300.00
.
const spookyItems = ['π»', 'π', 'πΈ'];
({ item: spookyItems[3] } = { item: 'π' });
console.log(spookyItems);
- A:
["π»", "π", "πΈ"]
- B:
["π»", "π", "πΈ", "π"]
- C:
["π»", "π", "πΈ", { item: "π" }]
- D:
["π»", "π", "πΈ", "[object Object]"]
Answer
By destructuring objects, we can unpack values from the right-hand object, and assign the unpacked value to the value of the same property name on the left-hand object. In this case, we're assigning the value "π" to spookyItems[3]
. This means that we're modifying the spookyItems
array, we're adding the "π" to it. When logging spookyItems
, ["π»", "π", "πΈ", "π"]
gets logged.
const name = 'Lydia Hallie';
const age = 21;
console.log(Number.isNaN(name));
console.log(Number.isNaN(age));
console.log(isNaN(name));
console.log(isNaN(age));
- A:
true
false
true
false
- B:
true
false
false
false
- C:
false
false
true
false
- D:
false
true
false
true
Answer
With the Number.isNaN
method, you can check if the value you pass is a numeric value and equal to NaN
. name
is not a numeric value, so Number.isNaN(name)
returns false
. age
is a numeric value, but is not equal to NaN
, so Number.isNaN(age)
returns false
.
With the isNaN
method, you can check if the value you pass is not a number. name
is not a number, so isNaN(name)
returns true. age
is a number, so isNaN(age)
returns false
.
const randomValue = 21;
function getInfo() {
console.log(typeof randomValue);
const randomValue = 'Lydia Hallie';
}
getInfo();
- A:
"number"
- B:
"string"
- C:
undefined
- D:
ReferenceError
Answer
Variables declared with the const
keyword are not referencable before their initialization: this is called the temporal dead zone. In the getInfo
function, the variable randomValue
is scoped in the functional scope of getInfo
. On the line where we want to log the value of typeof randomValue
, the variable randomValue
isn't initialized yet: a ReferenceError
gets thrown! The engine didn't go down the scope chain since we declared the variable randomValue
in the getInfo
function.
const myPromise = Promise.resolve('Woah some cool data');
(async () => {
try {
console.log(await myPromise);
} catch {
throw new Error(`Oops didn't work`);
} finally {
console.log('Oh finally!');
}
})();
- A:
Woah some cool data
- B:
Oh finally!
- C:
Woah some cool data
Oh finally!
- D:
Oops didn't work
Oh finally!
Answer
In the try
block, we're logging the awaited value of the myPromise
variable: "Woah some cool data"
. Since no errors were thrown in the try
block, the code in the catch
block doesn't run. The code in the finally
block always runs, "Oh finally!"
gets logged.
const emojis = ['π₯', ['β¨', 'β¨', ['π', 'π']]];
console.log(emojis.flat(1));
- A:
['π₯', ['β¨', 'β¨', ['π', 'π']]]
- B:
['π₯', 'β¨', 'β¨', ['π', 'π']]
- C:
['π₯', ['β¨', 'β¨', 'π', 'π']]
- D:
['π₯', 'β¨', 'β¨', 'π', 'π']
Answer
With the flat
method, we can create a new, flattened array. The depth of the flattened array depends on the value that we pass. In this case, we passed the value 1
(which we didn't have to, that's the default value), meaning that only the arrays on the first depth will be concatenated. ['π₯']
and ['β¨', 'β¨', ['π', 'π']]
in this case. Concatenating these two arrays results in ['π₯', 'β¨', 'β¨', ['π', 'π']]
.
class Counter {
constructor() {
this.count = 0;
}
increment() {
this.count++;
}
}
const counterOne = new Counter();
counterOne.increment();
counterOne.increment();
const counterTwo = counterOne;
counterTwo.increment();
console.log(counterOne.count);
- A:
0
- B:
1
- C:
2
- D:
3
Answer
counterOne
is an instance of the Counter
class. The counter class contains a count
property on its constructor, and an increment
method. First, we invoked the increment
method twice by calling counterOne.increment()
. Currently, counterOne.count
is 2
.
Then, we create a new variable counterTwo
, and set it equal to counterOne
. Since objects interact by reference, we're just creating a new reference to the same spot in memory that counterOne
points to. Since it has the same spot in memory, any changes made to the object that counterTwo
has a reference to, also apply to counterOne
. Currently, counterTwo.count
is 2
.
We invoke the counterTwo.increment()
, which sets the count
to 3
. Then, we log the count on counterOne
, which logs 3
.
const myPromise = Promise.resolve(Promise.resolve('Promise!'));
function funcOne() {
myPromise.then(res => res).then(res => console.log(res));
setTimeout(() => console.log('Timeout!', 0));
console.log('Last line!');
}
async function funcTwo() {
const res = await myPromise;
console.log(await res);
setTimeout(() => console.log('Timeout!', 0));
console.log('Last line!');
}
funcOne();
funcTwo();
- A:
Promise! Last line! Promise! Last line! Last line! Promise!
- B:
Last line! Timeout! Promise! Last line! Timeout! Promise!
- C:
Promise! Last line! Last line! Promise! Timeout! Timeout!
- D:
Last line! Promise! Promise! Last line! Timeout! Timeout!
Answer
First, we invoke funcOne
. On the first line of funcOne
, we call the myPromise
promise, which is an asynchronous operation. While the engine is busy completing the promise, it keeps on running the function funcOne
. The next line is the asynchronous setTimeout
function, from which the callback is sent to the Web API. (see my article on the event loop here.)
Both the promise and the timeout are asynchronous operations, the function keeps on running while it's busy completing the promise and handling the setTimeout
callback. This means that Last line!
gets logged first, since this is not an asynchonous operation. This is the last line of funcOne
, the promise resolved, and Promise!
gets logged. However, since we're invoking funcTwo()
, the call stack isn't empty, and the callback of the setTimeout
function cannot get added to the callstack yet.
In funcTwo
we're, first awaiting the myPromise promise. With the await
keyword, we pause the execution of the function until the promise has resolved (or rejected). Then, we log the awaited value of res
(since the promise itself returns a promise). This logs Promise!
.
The next line is the asynchronous setTimeout
function, from which the callback is sent to the Web API.
We get to the last line of funcTwo
, which logs Last line!
to the console. Now, since funcTwo
popped off the call stack, the call stack is empty. The callbacks waiting in the queue (() => console.log("Timeout!")
from funcOne
, and () => console.log("Timeout!")
from funcTwo
) get added to the call stack one by one. The first callback logs Timeout!
, and gets popped off the stack. Then, the second callback logs Timeout!
, and gets popped off the stack. This logs Last line! Promise! Promise! Last line! Timeout! Timeout!
// sum.js
export default function sum(x) {
return x + x;
}
// index.js
import * as sum from './sum';
- A:
sum(4)
- B:
sum.sum(4)
- C:
sum.default(4)
- D: Default aren't imported with
*
, only named exports
Answer
With the asterisk *
, we import all exported values from that file, both default and named. If we had the following file:
// info.js
export const name = 'Lydia';
export const age = 21;
export default 'I love JavaScript';
// index.js
import * as info from './info';
console.log(info);
The following would get logged:
{
default: "I love JavaScript",
name: "Lydia",
age: 21
}
For the sum
example, it means that the imported value sum
looks like this:
{ default: function sum(x) { return x + x } }
We can invoke this function, by calling sum.default
const handler = {
set: () => console.log('Added a new property!'),
get: () => console.log('Accessed a property!'),
};
const person = new Proxy({}, handler);
person.name = 'Lydia';
person.name;
- A:
Added a new property!
- B:
Accessed a property!
- C:
Added a new property!
Accessed a property!
- D: Nothing gets logged
Answer
With a Proxy object, we can add custom behavior to an object that we pass to it as the second argument. In tis case, we pass the handler
object which contained to properties: set
and get
. set
gets invoked whenever we set property values, get
gets invoked whenever we get (access) property values.
The first argument is an empty object {}
, which is the value of person
. To this object, the custom behavior specified in the handler
object gets added. If we add a property to the person
object, set
will get invoked. If we access a property on the person
object, get
gets invoked.
First, we added a new property name
to the proxy object (person.name = "Lydia"
). set
gets invoked, and logs "Added a new property!"
.
Then, we access a property value on the proxy object, the get
property on the handler object got invoked. "Accessed a property!"
gets logged.
const person = { name: 'Lydia Hallie' };
Object.seal(person);
- A:
person.name = "Evan Bacon"
- B:
person.age = 21
- C:
delete person.name
- D:
Object.assign(person, { age: 21 })
Answer
With Object.seal
we can prevent new properies from being added, or existing properties to be removed.
However, you can still modify the value of existing properties.
const person = {
name: 'Lydia Hallie',
address: {
street: '100 Main St',
},
};
Object.freeze(person);
- A:
person.name = "Evan Bacon"
- B:
delete person.address
- C:
person.address.street = "101 Main St"
- D:
person.pet = { name: "Mara" }
Answer
The Object.freeze
method freezes an object. No properties can be added, modified, or removed.
However, it only shallowly freezes the object, meaning that only direct properties on the object are frozen. If the property is another object, like address
in this case, the properties on that object aren't frozen, and can be modified.
const add = x => x + x;
function myFunc(num = 2, value = add(num)) {
console.log(num, value);
}
myFunc();
myFunc(3);
- A:
2
4
and3
6
- B:
2
NaN
and3
NaN
- C:
2
Error
and3
6
- D:
2
4
and3
Error
Answer
First, we invoked myFunc()
without passing any arguments. Since we didn't pass arguments, num
and value
got their default values: num is 2
, and value
the returned value of the function add
. To the add
function, we pass num
as an argument, which had the value of 2
. add
returns 4
, which is the value of value
.
Then, we invoked myFunc(3)
and passed the value 3
as the value for the argument num
. We didn't pass an argument for value
. Since we didn't pass a value for the value
argument, it got the default value: the returned value of the add
function. To add
, we pass num
, which has the value of 3
. add
returns 6
, which is the value of value
.
class Counter {
#number = 10
increment() {
this.#number++
}
getNum() {
return this.#number
}
}
const counter = new Counter()
counter.increment()
console.log(counter.#number)
- A:
10
- B:
11
- C:
undefined
- D:
SyntaxError
Answer
In ES2020, we can add private variables in classes by using the #
. We cannot access these variables outside of the class. When we try to log counter.#number
, a SyntaxError gets thrown: we cannot acccess it outside the Counter
class!
const teams = [
{ name: 'Team 1', members: ['Paul', 'Lisa'] },
{ name: 'Team 2', members: ['Laura', 'Tim'] },
];
function* getMembers(members) {
for (let i = 0; i < members.length; i++) {
yield members[i];
}
}
function* getTeams(teams) {
for (let i = 0; i < teams.length; i++) {
// β¨ SOMETHING IS MISSING HERE β¨
}
}
const obj = getTeams(teams);
obj.next(); // { value: "Paul", done: false }
obj.next(); // { value: "Lisa", done: false }
- A:
yield getMembers(teams[i].members)
- B:
yield* getMembers(teams[i].members)
- C:
return getMembers(teams[i].members)
- D:
return yield getMembers(teams[i].members)
Answer
In order to iterate over the members
in each element in the teams
array, we need to pass teams[i].members
to the getMembers
generator function. The generator function returns a generator object. In order to iterate over each element in this generator object, we need to use yield*
.
If we would've written yield
, return yield
, or return
, the entire generator function would've gotten returned the first time we called the next
method.
const person = {
name: 'Lydia Hallie',
hobbies: ['coding'],
};
function addHobby(hobby, hobbies = person.hobbies) {
hobbies.push(hobby);
return hobbies;
}
addHobby('running', []);
addHobby('dancing');
addHobby('baking', person.hobbies);
console.log(person.hobbies);
- A:
["coding"]
- B:
["coding", "dancing"]
- C:
["coding", "dancing", "baking"]
- D:
["coding", "running", "dancing", "baking"]
Answer
The addHobby
function receives two arguments, hobby
and hobbies
with the default value of the hobbies
array on the person
object.
First, we invoke the addHobby
function, and pass "running"
as the value for hobby
and an empty array as the value for hobbies
. Since we pass an empty array as the value for y
, "running"
gets added to this empty array.
Then, we invoke the addHobby
function, and pass "dancing"
as the value for hobby
. We didn't pass a value for hobbies
, so it gets the default value, the hobbies
property on the person
object. We push the hobby dancing
to the person.hobbies
array.
Last, we invoke the addHobby
function, and pass "bdaking"
as the value for hobby
, and the person.hobbies
array as the value for hobbies
. We push the hobby baking
to the person.hobbies
array.
After pushing dancing
and baking
, the value of person.hobbies
is ["coding", "dancing", "baking"]
class Bird {
constructor() {
console.log("I'm a bird. π¦’");
}
}
class Flamingo extends Bird {
constructor() {
console.log("I'm pink. πΈ");
super();
}
}
const pet = new Flamingo();
- A:
I'm pink. πΈ
- B:
I'm pink. πΈ
I'm a bird. π¦’
- C:
I'm a bird. π¦’
I'm pink. πΈ
- D: Nothing, we didn't call any method
Answer
We create the variable pet
which is an instance of the Flamingo
class. When we instantiate this instance, the constructor
on Flamingo
gets called. First, "I'm pink. πΈ"
gets logged, after which we call super()
. super()
calls the constructor of the parent class, Bird
. THe constructor in Bird
gets called, and logs "I'm a bird. π¦’"
.
const emojis = ['π', 'π
πΌ', 'π', 'β'];
/* 1 */ emojis.push('π¦');
/* 2 */ emojis.splice(0, 2);
/* 3 */ emojis = [...emojis, 'π₯'];
/* 4 */ emojis.length = 0;
- A: 1
- B: 1 and 2
- C: 3 and 4
- D: 3
Answer
The const
keyword simply means we cannot redeclare the value of that variable, it's read-only. However, the value itself isn't immutable. The propeties on the emojis
array can be modified, for example by pushing new values, splicing them, or setting the length of the array to 0.
144. What do we need to add to the person
object to get ["Lydia Hallie", 21]
as the output of [...person]
?
const person = {
name: "Lydia Hallie",
age: 21
}
[...person] // ["Lydia Hallie", 21]
- A: Nothing, object are iterable by default
- B:
*[Symbol.iterator]() { for (let x in this) yield* this[x] }
- C:
*[Symbol.iterator]() { yield* Object.values(this) }
- D:
*[Symbol.iterator]() { for (let x in this) yield this }
Answer
Objects aren't iterable by default. An iterable is an iterable if the iterator protocol is present. We can add this manually by adding the iterator symbol [Symbol.iterator]
, which has to return a generator object, for example by making it a generator function *[Symbol.iterator]() {}
. This generator function has to yield the Object.values
of the person
object if we want it to return the array ["Lydia Hallie", 21]
: yield* Object.values(this)
.
let count = 0;
const nums = [0, 1, 2, 3];
nums.forEach(num => {
if (num) count += 1
})
console.log(count)
- A: 1
- B: 2
- C: 3
- D: 4
Answer
The if
condition within the forEach
loop checks whether the value of num
is truthy or falsy. Since the first number in the nums
array is 0
, a falsy value, the if
statement's code block won't be executed. count
only gets incremented for the other 3 numbers in the nums
array, 1
, 2
and 3
. Since count
gets incremented by 1
3 times, the value of count
is 3
.
function getFruit(fruits) {
console.log(fruits?.[1]?.[1])
}
getFruit([['π', 'π'], ['π']])
getFruit()
getFruit([['π'], ['π', 'π']])
- A:
null
,undefined
, π - B:
[]
,null
, π - C:
[]
,[]
, π - D:
undefined
,undefined
, π
Answer
The ?
allows us to optionally access deeper nested properties within objects. We're trying to log the item on index 1
within the subarray that's on index 1
of the fruits
array. If the subarray on index 1
in the fruits
array doesn't exist, it'll simply return undefined
. If the subarray on index 1
in the fruits
array exists, but this subarray doesn't have an item on its 1
index, it'll also return undefined
.
First, we're trying to log the second item in the ['π']
subarray of [['π', 'π'], ['π']]]
. This subarray only contains one item, which means there is no item on index 1
, and returns undefined
.
Then, we're invoking the getFruits
function without passing a value as an argument, which means that fruits
has a value of undefined
by default. Since we're conditionally chaining the item on index 1
offruits
, it returns undefined
since this item on index 1
does not exist.
Lastly, we're trying to log the second item in the ['π', 'π']
subarray of ['π'], ['π', 'π']
. The item on index 1
within this subarray is π
, which gets logged.
class Calc {
constructor() {
this.count = 0
}
increase() {
this.count ++
}
}
const calc = new Calc()
new Calc().increase()
console.log(calc.count)
- A:
0
- B:
1
- C:
undefined
- D:
ReferenceError
Answer
We set the variable calc
equal to a new instance of the Calc
class. Then, we instantiate a new instance of Calc
, and invoke the increase
method on this instance. Since the count property is within the constructor of the Calc
class, the count property is not shared on the prototype of Calc
. This means that the value of count has not been updated for the instance calc points to, count is still 0
.
const user = {
email: "[email protected]",
password: "12345"
}
const updateUser = ({ email, password }) => {
if (email) {
Object.assign(user, { email })
}
if (password) {
user.password = password
}
return user
}
const updatedUser = updateUser({ email: "[email protected]" })
console.log(updatedUser === user)
- A:
false
- B:
true
- C:
TypeError
- D:
ReferenceError
Answer
The updateUser
function updates the values of the email
and password
properties on user, if their values are passed to the function, after which the function returns the user
object. The returned value of the updateUser
function is the user
object, which means that the value of updatedUser is a reference to the same user
object that user
points to. updatedUser === user
equals true
.
const fruit = ['π', 'π', 'π']
fruit.slice(0, 1)
fruit.splice(0, 1)
fruit.unshift('π')
- A:
['π', 'π', 'π']
- B:
['π', 'π']
- C:
['π', 'π', 'π']
- D:
['π', 'π', 'π', 'π']
Answer
First, we invoke the slice
method on the fruit array. The slice method does not modify the original array, but returns the value that it sliced off the array: the banana emoji.
Then, we invoke the splice
method on the fruit array. The splice method does modify the original array, which means that the fruit array now consists of ['π', 'π']
.
At last, we invoke the unshift
method on the fruit
array, which modifies the original array by adding the provided value, βπβ in this case, as the first element in the array. The fruit array now consists of ['π', 'π', 'π']
.
const animals = {};
let dog = { emoji: 'πΆ' }
let cat = { emoji: 'π' }
animals[dog] = { ...dog, name: "Mara" }
animals[cat] = { ...cat, name: "Sara" }
console.log(animals[dog])
- A:
{ emoji: "πΆ", name: "Mara" }
- B:
{ emoji: "π", name: "Sara" }
- C:
undefined
- D:
ReferenceError
Answer
Object keys are converted to strings.
Since the value of dog
is an object, animals[dog]
actually means that weβre creating a new property called "object Object"
equal to the new object. animals["object Object"]
is now equal to { emoji: "πΆ", name: "Mara"}
.
cat
is also an object, which means that animals[cat]
actually means that weβre overwriting the value of animals[``"``object Object``"``]
with the new cat properties.
Logging animals[dog]
, or actually animals["object Object"]
since converting the dog
object to a string results "object Object"
, returns the { emoji: "π", name: "Sara" }
.
const user = {
email: "[email protected]",
updateEmail: email => {
this.email = email
}
}
user.updateEmail("[email protected]")
console.log(user.email)
- A:
[email protected]
- B:
[email protected]
- C:
undefined
- D:
ReferenceError
Answer
The updateEmail
function is an arrow function, and is not bound to the user
object. This means that the this
keyword is not referring to the user
object, but refers to the global scope in this case. The value of email
within the user
object does not get updated. When logging the value of user.email
, the original value of [email protected]
gets returned.
const promise1 = Promise.resolve('First')
const promise2 = Promise.resolve('Second')
const promise3 = Promise.reject('Third')
const promise4 = Promise.resolve('Fourth')
const runPromises = async () => {
const res1 = await Promise.all([promise1, promise2])
const res2 = await Promise.all([promise3, promise4])
return [res1, res2]
}
runPromises()
.then(res => console.log(res))
.catch(err => console.log(err))
- A:
[['First', 'Second'], ['Fourth']]
- B:
[['First', 'Second'], ['Third', 'Fourth']]
- C:
[['First', 'Second']]
- D:
'Third'
Answer
The Promise.all
method runs the passed promises in parallel. If one promise fails, the Promise.all
method _rejects) with the value of the rejected promise. In this case, promise3
rejected with the value "Third"
. Weβre catching the rejected value in the chained catch
method on the runPromises
invocation to catch any errors within the runPromises
function. Only "Third"
gets logged, since promise3
rejected with this value.
const keys = ["name", "age"]
const values = ["Lydia", 22]
const method = /* ?? */
Object[method](keys.map((_, i) => {
return [keys[i], values[i]]
})) // { name: "Lydia", age: 22 }
- A:
entries
- B:
values
- C:
fromEntries
- D:
forEach
Answer
The fromEntries
method turns a 2d array into an object. The first element in each subarray will be the key, and the second element in each subarray will be the value. In this case, weβre mapping over the keys
array, which returns an array which first element is the item on the key array on the current index, and the second element is the item of the values array on the current index.
This creates an array of subarrays containing the correct keys and values, which results in { name: "Lydia", age: 22 }
const createMember = ({ email, address = {}}) => {
const validEmail = /.+\@.+\..+/.test(email)
if (!validEmail) throw new Error("Valid email pls")
return {
email,
address: address ? address : null
}
}
const member = createMember({ email: "[email protected]" })
console.log(member)
- A:
{ email: "[email protected]", address: null }
- B:
{ email: "[email protected]" }
- C:
{ email: "[email protected]", address: {} }
- D:
{ email: "[email protected]", address: undefined }
Answer
The default value of address
is an empty object {}
. When we set the variable member
equal to the object returned by the createMember
function, we didn't pass a value for address, which means that the value of address is the default empty object {}
. An empty object is a truthy value, which means that the condition of the address ? address : null
conditional returns true
. The value of address is the empty object {}
.
let randomValue = { name: "Lydia" }
randomValue = 23
if (!typeof randomValue === "string") {
console.log("It's not a string!")
} else {
console.log("Yay it's a string!")
}
- A:
It's not a string!
- B:
Yay it's a string!
- C:
TypeError
- D:
undefined
Answer
The condition within the if
statement checks whether the value of !typeof randomValue
is equal to "string"
. The !
operator converts the value to a boolean value. If the value is truthy, the returned value will be false
, if the value is falsy, the returned value will be true
. In this case, the returned value of typeof randomValue
is the truthy value "string"
, meaning that the value of !typeof randomValue
is the boolean value false
.
!typeof randomValue === "string"
always returns false, since we're actually checking false === "string"
. Since the condition returned false
, the code block of the else
statement gets run, and Yay it's a string!
gets logged.