Because pointers can be ugh
To understand a pointer, let's review "regular" variables first. If you're familiar with a programming language without pointers like JavaScript, this is what you think when you hear "variable".
When declaring a variable by identifier (or name), the variable is synonymous with its value.
int number = 3;
std::cout << "number is initialized with a value of " << number << "\n";
// Outputs: number is initialized with a value of 3The value of the variable can be modified directly.
number++;
number = number * 2;
std::cout << "After modifying number, its value is now " << number << "\n";
// Outputs: After modifying number, its value is now 8Every value lives at memory addresses. The memory address of a variable can be accessed using the & operator.
std::cout << "The number variable's value lives at memory address " << &number << "\n";A pointer is a variable that holds a memory address where a value lives. [ need a super brief description of why pointers are useful here ]
A pointer is a variable that holds a memory address where a value lives.
A pointer is declared using the * operator before an identifier.
As C++ is a statically typed language, the type is required to declare a pointer. This is the type of data that will live at the memory address it points to.
int *numberPointer;
std::cout << "The numberPointer points to the memory address " << numberPointer << ".\n";
// Outputs: The numberPointer points to the memory address 0x0We've initialized a pointer, but it points nowhere, it has no memory address. Note that 0x0 is equivalent to 0. It's just the hexadecimal representation of 0, which memory addresses are expressed in.
We can set the memory address the pointer points to using the memory address of the number variable.
The memory address of any variable can be accessed using the & operator.
numberPointer = &number;
std::cout << "The numberPointer points to the memory address `" << numberPointer << "`, which is the memory address of number\n";
// Outputs: The numberPointer points to the memory address 0x1001054a0
// NOTE: if you run this code, you will get a different memory address.When declaring a pointer, we can alternatively instantiate a variable at its memory address.
int *numberPointer = new int(3);Instantiating a variable with the new operator always returns a pointer.
The value at the memory address pointed to can be accessed with the dereference operator, *.
*numberPointer = 4;
std::cout << "The variable's value has been modified using the pointer. The variable's value is now " << number << "\n";
// Outputs: The variable's value has been modified using the pointer. The variable's value is now 4We've shown that the asterisk operator * has two semantic meanings: to declare a pointer variable (i.e. int *numberPointer) and also to access the value a pointer points to (*numberPointer = 4). This is confusing but just part of the syntax we need to internalize.
In C++, variables are passed to a function by value. When calling a function with an argument, a new variable is instantiated internally and assigned the value passed in the function call.
Any modifications to the value inside the function are performed to this new variable; the variable that was invoked with the function call is unchanged.
void setValueToFive(int x)
{
std::cout << &x << "\n";
x = 5;
}
int x = 3;
setValueToFive( x );
std::cout << "The value of x is " << x << ", setValueToFive() did not modify x.\n";
// Outputs: The value of x is 3, setValueToFive() did not modify xA function can be written to perform the same task but instead accept a pointer as the argument. This lowers the memory footprint of the program. Unnecessary duplicate variables aren't created. The function can modify the variable's value directly. Any modifications to the variable in the function affect the variable here in this scope too.
void setPointerValueToFive(int *x)
{
*x = 5;
}
int x = 3;
setPointerValueToFive( &x );
std::cout << "The value of x is " << x << ", setPointerValueToFive() modified x.\n";
// Outputs: The value of x is 5, setPointerValueToFive() modified xIn C++ (not in C), you can pass a variable by reference by passing in a variable,
and defining the function's parameter like int &x. Any modifications to the variable inside the function will also affect the variable here in this scope too.
void setValueToFiveWithReference(int& x)
{
x = 5;
}
int x = 3;
setValueToFiveWithReference( x );
std::cout << "The value of x is " << x << ", setPointerValueToFive() modified x.\n";
// Outputs: The value of x is 5, setPointerValueToFive() modified xWhich to use is a matter of style, and whether you want to perform pointer arithmetic in the function. Google has specific guidelines for which to use where.
The pointer is a variable in itself, and all variables are stored somewhere in memory.
We can see the memory address the pointer's memory address is stored.
In response to the above comment - the new link is at https://google.github.io/styleguide/cppguide.html#Reference_Arguments now it seems!