What will the code below output? Explain your answer.
console.log(0.1 + 0.2);
console.log(0.1 + 0.2 == 0.3);Answer
An educated answer to this question would simply be: “You can’t be sure. it might print out 0.3 and true, or it might not. Numbers in JavaScript are all treated with floating point precision, and as such, may not always yield the expected results.”
The example provided above is classic case that demonstrates this issue. Surprisingly, it will print out:
0.30000000000000004
falseConsider the two functions below. Will they both return the same thing? Why or why not?
function foo1()
{
return {
bar: "hello"
};
}
function foo2()
{
return
{
bar: "hello"
};
}Answer
Surprisingly, these two functions will not return the same thing. Rather:
console.log("foo1 returns:");
console.log(foo1());
console.log("foo2 returns:");
console.log(foo2());will yield:
foo1 returns:
Object {bar: "hello"}
foo2 returns:
undefined Not only is this surprising, but what makes this particularly gnarly is that foo2() returns undefined without any error being thrown.
The reason for this has to do with the fact that semicolons are technically optional in JavaScript (although omitting them is generally really bad form). As a result, when the line containing the return statement (with nothing else on the line) is encountered in foo2(), a semicolon is automatically inserted immediately after the return statement.
Write a sum method which will work properly when invoked using either syntax below.
console.log(sum(2,3)); // Outputs 5
console.log(sum(2)(3)); // Outputs 5Answer
There are (at least) two ways to do this:
METHOD 1
function sum(x) {
if (arguments.length == 2) {
return arguments[0] + arguments[1];
} else {
return function(y) { return x + y; };
}
}In JavaScript, functions provide access to an arguments object which provides access to the actual arguments passed to a function. This enables us to use the length property to determine at runtime the number of arguments passed to the function.
If two arguments are passed, we simply add them together and return.
Otherwise, we assume it was called in the form sum(2)(3), so we return an anonymous function that adds together the argument passed to sum() (in this case 2) and the argument passed to the anonymous function (in this case 3).
METHOD 2
function sum(x, y) {
if (y !== undefined) {
return x + y;
} else {
return function(y) { return x + y; };
}
}When a function is invoked, JavaScript does not require the number of arguments to match the number of arguments in the function definition. If the number of arguments passed exceeds the number of arguments in the function definition, the excess arguments will simply be ignored. On the other hand, if the number of arguments passed is less than the number of arguments in the function definition, the missing arguments will have a value of undefined when referenced within the function. So, in the above example, by simply checking if the 2nd argument is undefined, we can determine which way the function was invoked and proceed accordingly.
What will the code below output to the console and why?
(function(){
var a = b = 3;
})();
console.log("a defined? " + (typeof a !== 'undefined'));
console.log("b defined? " + (typeof b !== 'undefined'));Answer
Since both a and b are defined within the enclosing scope of the function, and since the line they are on begins with the var keyword, most JavaScript developers would expect typeof a and typeof b to both be undefined in the above example.
However, that is not the case. The issue here is that most developers incorrectly understand the statement var a = b = 3; to be shorthand for:
var b = 3;
var a = b;But in fact, var a = b = 3; is actually shorthand for:
b = 3;
var a = b;As a result (if you are not using strict mode), the output of the code snippet would be:
a defined? false
b defined? trueBut how can b be defined outside of the scope of the enclosing function? Well, since the statement var a = b = 3; is shorthand for the statements b = 3; and var a = b;, b ends up being a global variable (since it is not preceded by the var keyword) and is therefore still in scope even outside of the enclosing function.
Note that, in strict mode (i.e., with use strict), the statement var a = b = 3; will generate a runtime error of ReferenceError: b is not defined, thereby avoiding any headfakes/bugs that might othewise result. (Yet another prime example of why you should use use strict as a matter of course in your code!)
In what order will the numbers 1-4 be logged to the console when the code below is executed? Why?
(function() {
console.log(1);
setTimeout(function(){console.log(2)}, 1000);
setTimeout(function(){console.log(3)}, 0);
console.log(4);
})();Answer
The values will be logged in the following order:
1
4
3
2Let’s first explain the parts of this that are presumably more obvious:
1 and 4 are displayed first since they are logged by simple calls to console.log() without any delay
2 is displayed after 3 because 2 is being logged after a delay of 1000 msecs (i.e., 1 second) whereas 3 is being logged after a delay of 0 msecs.
OK, fine. But if 3 is being logged after a delay of 0 msecs, doesn’t that mean that it is being logged right away? And, if so, shouldn’t it be logged before 4, since 4 is being logged by a later line of code?
The answer has to do with properly understanding JavaScript events and timing.
The browser has an event loop which checks the event queue and processes pending events. For example, if an event happens in the background (e.g., a script onload event) while the browser is busy (e.g., processing an onclick), the event gets appended to the queue. When the onclick handler is complete, the queue is checked and the event is then handled (e.g., the onload script is executed).
Similarly, setTimeout() also puts execution of its referenced function into the event queue if the browser is busy.
When a value of zero is passed as the second argument to setTimeout(), it attempts to execute the specified function “as soon as possible”. Specifically, execution of the function is placed on the event queue to occur on the next timer tick. Note, though, that this is not immediate; the function is not executed until the next tick. That’s why in the above example, the call to console.log(4) occurs before the call to console.log(3) (since the call to console.log(3) is invoked via setTimeout, so it is slightly delayed).
What are the differences between ES6 class and ES5 function constructors?
// ES5 Function Constructor
function Person(name) {
this.name = name;
}
// ES6 Class
class Person {
constructor(name) {
this.name = name;
}
}Answer
The main difference in the constructor comes when using inheritance. If we want to create a Student class that subclasses Person and add a studentId field, this is what we have to do in addition to the above:
// ES5 Function Constructor
function Student(name, studentId) {
// Call constructor of superclass to initialize superclass-derived members.
Person.call(this, name);
// Initialize subclass's own members.
this.studentId = studentId;
}
Student.prototype = Object.create(Person.prototype);
Student.prototype.constructor = Student;
// ES6 Class
class Student extends Person {
constructor(name, studentId) {
super(name);
this.studentId = studentId;
}
}It's much more verbose to use inheritance in ES5 and the ES6 version is easier to understand and remember.
What is IIFEs (Immediately Invoked Function Expressions)?
Answer
It’s an Immediately-Invoked Function Expression, or IIFE for short. It executes immediately after it’s created:
(function IIFE(){
console.log( "Hello!" );
})();
// "Hello!"This pattern is often used when trying to avoid polluting the global namespace, because all the variables used inside the IIFE (like in any other normal function) are not visible outside its scope.