Difference between &a, *(&a) and *(*(&a))-CodePudding

Code

#include <iostream>

using namespace std;

int main()
{
    int i = 7;
    int *a = &i;
    cout << &a << endl; //This Line
    cout << *(&a) << endl; //This Line
    cout << *(*(&a)) << endl; //This Line
}

Outputs

0x6a12fffdf8
0x6a12fffe00
7

Hi all. I am not very sure what the difference between these three lines is (see code comments). I know the first line prints the address, but I am not sure about the other two.

CodePudding user response：

Lets go line by line:

int i = 7;

i is an int initialized with the integer literal 7.

int *a = &i;

a is a pointer, int*, and is initialized with the address of i. & is the addressof operator here.

cout << &a << endl; //This Line

a itself is also stored at some memory address. The addressof operator applied to a yields that memory address.

cout << *(&a) << endl; //This Line

Here * is the dereference operator. &, addressof, and *, dereference, cancel each other. This line prints the value of a, the address of i.

cout << *(*(&a)) << endl; //This Line

We already know that *(&a) is the value of a, the memory address of i. Applying the dereference operator to the address of i yields the value of i.

In memory it looks like this:

 address: &a             &i == a
          0x6a12fffdf8   0x6a12fffe00

 value:   a              i == *a
          0x6a12fffe00   7

CodePudding user response：

You have some data, int i = 7;. This integer is stored in memory (specifically, the stack).

You have a pointer, a, that points to (addresses) this memory int *a = &i; because the int * datatype means an int pointer, and &i means the "address of i".

Your pointer is also stored in memory (again, the stack). You can get its address in much the same way, using &a.

cout << &a << endl; therefore prints the address of the pointer, a.

To dereference (get the value of) a pointer, you use *.

cout << *(&a) << endl; therefore prints the dereferenced value of the address of a, which is the value of a (and therefore the address of i).

Your final line cout << *(*(&a)) << endl; reads as

"Print the dereferenced value of the dereferenced value of the address of a", which is simply i.

CodePudding user response：

Case 1

Here we consider the expression:

&a  //this gives us the address of the variable `a`

The above expression gives us the address of variable a.

Case 2

Here we consider the expression:

*(&a)  //this gives us the data to which `&a` points which is nothing but the address of variable i

The above expression gives us the data to which &a points. But from case 1 we already know that &a is nothing but the address of variable a. Thus, *(&a) is the same as the address of variable i.

Case 3

Here we consider the expression:

*(*(&a))

The above expression gives us the data to which *(&a) points. Since we already know from case 2 that *(&a) gives us the address of variable i, this implies that *(*(&a)) gives us i itself.