Pointers and addresses in C-CodePudding

I am messing around a little with pointers. Please take a look at the following results (addresses).

1st code:

#include <iostream>

int main(){
    int a = 5;
    void* pointer = &a;
    std::cout << a << std::endl << &a << std::endl;
    std::cout << pointer << std::endl ;
    std::cin.get();
}

Result:

2nd code:

#include <iostream>

int main(){
    int a = 5;
    void* pointer = &a;
    std::cout << a << std::endl << &a << std::endl;
    std::cout << &pointer << std::endl ;
    std::cin.get();
}

Result:

Why does the address of the variable a change between the two codes?

CodePudding user response：

In the first case, you never take the address of pointer, so pointer can be stored in a register, or even not at all. (Modern compilers are very clever, and modern machines have many registers.)

For instance, gcc 11 keeps the value in a register without optimization, and with -O2 it just inserts the address of a directly. (Assembly here, for the curious.)

In the second case, you do take the address of pointer, so it must be stored somewhere in memory.
This means that a might be stored in a different place in order to make room for it.

Also, some platforms randomize storage locations in order to make programs less hackable, so it's usually not a good idea to assume that things will have the same address in different "runs".

CodePudding user response：

In the first code poiner stores the address of variable a, and by command.

std::cout<<pointer<<std::endl;

You print the address of a. That's it. In the second code pointer also stores the address of variable a, but &pointer is the address of variable pointer. Try the following

#include <iostream>
int main(){
    int a = 5;
    void* pointer = &a;
    void** pointer_to_pointer = &pointer;
    std::cout << a << std::endl << &a << std::endl << pointer << std::endl;
    std::cout << &pointer << std::endl << pointer_to_pointer;
    std::cin.get();
}

The output for me is

5
0x7aba1bd92e94
0x7aba1bd92e94
0x7aba1bd92e98
0x7aba1bd92e98

CodePudding user response：

It is very simple. int a and void* pointer are two distinct variables or I better say memory locations on the stack. a holds a value like 5 in its location. pointer holds the address to a's memory location. pointer itself is stored in a different location and when you write std::cout << &pointer << std::endl; it will print the address of the pointer variable, not the contents of it which is a's address.

As a simplified example:

Consider 0x4 as the address of pointer itself and the value inside it is 0xC. This value points to a's location. In order to read the value of a(which is 5), you first have to go to 0x4 to read its content. Its content is 0xC and now you have successfully found out that a's location is 0xC. Then you have to go to 0xC and at that address, you will find the value 5.

You look at -> 0x4( content == 0xC ) -> 0xC( content == 5 ) -> done!