Pointer to an argument passed as value?

Question

In the example below:

#include <stdio.h>
#include <stdlib.h>

void test1 (int x)
{
    int * xp = &x;
    printf("Address of xp (test1): [%p]\n", &xp);
    printf("xp points to: [%p] which is the same as [%p]\n", xp, &x);
}

int main()
{
    int x = 4;
    printf("Address of x (main): [%p] \n", &x);
    test1(x);
    return 0;
}

Output:

Address of x (main): [0xbfca0c5c] 
Address of xp (test1): [0xbfca0c2c] 
xp points to: [0xbfca0c40] which is the same as [0xbfca0c40] 

• What happens to the address 0xbfca0c40 (address of the argument in test1) outside of the function?
• Where is it located in memory?

Answer 1

That is defined by the implementation.

It's very likely on the machine stack, since the stack is very common way to implement both argument-passing and local variables. The stack space is freed when the test1() function exits, so the memory in question can be re-used.

Answer 2

int x; in the test1 formal parameter list is a variable local to test1. It has similar status to if you also went int y; as a declaration inside test1. It has automatic storage duration.

After test1 exits, the variable no longer exists.

Typically, compilers implement automatic storage via a block of memory that contains a stack structure. When you call a function, new space is claimed on top of the stack for that function's automatic variables (and the return address, and any registers that need to be saved); and when you leave the function, the stack is "popped" back to where it was before you called the function.

Answer 3

1. The adress will still contain the value of x untill some program will reuse it.
2. The adress is on the stack, as the variable x is a function argument.