Why does this code give two different memory addresses?

[Swarvy]

I was just looking at this code and I can't explain why it's working in this way. I would expect the value given by &x and ptr to be the same, but it doesn't if you compile and run the program.

int main(int argc, char *argv[])
{
       int x=5, *ptr;           // x set to 5 for example only

       *ptr = x;

       printf("%x\n%x\n\n, ptr, &x);
       free(ptr);

       return 0;
}

I can't understand why they give different addresses, but for some reason, they do. I would be grateful if someone could explain to me why this is the case.

[laserlight]

Perhaps you would like to take a good look at the code below and observe the output:

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

int main(void)
{
    int x = 5;
    int *ptr;

    printf("&x = &#37;p\n", (void*)&x);

    ptr = &x;
    printf("ptr = %p\n", (void*)ptr);

    ptr = malloc(sizeof(*ptr));
    *ptr = x;
    printf("ptr = %p\n", (void*)ptr);
    free(ptr);

    return 0;
}

Note that pointers are printed with the %p format specifier, and should be pointers to void when printed, hence the cast.

EDIT:
Oops, I noticed a mistake: dynamic memory allocation is in fact needed, so the mistake is not with free(), but the lack of malloc().

[Elysia]

You are invoking undefined behavior because you are trying to use the pointer as a storage.
You aren't assigning an address to the pointer! How then, can they be the same?