Why does this code give two different memory addresses?

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.

Code:

---

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.

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

Code:

---

#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().

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?