question about a compiler warning

Here's a simple program that uses a pointer. The program compiles but issues this warning. The compiler I use is gcc.

gcc -g -ansi -pedantic -Wall ex9-2-4.c -o ex9-2-4
ex9-2-4.c: In function 'main':
ex9-2-4.c:12 warning: int format, pointer arg (arg 2)

Here's the code.

Code:

---

/* ex9-2-4.c */

#include <stdio.h>

int main(void)
{
        int cost, *p_cost;
        p_cost = &cost;

        cost = 100;

        printf("\nPointer value: %d, points at value: %d\n\n", p_cost, *p_cost);

        return 0;
}

---

What conversion specifier do I use for pointers?

P.S. If this question has been asked before on this message board, I'd appreciate someone giving me a link to the appropriate post. Thanks.

Code:

---

printf("\nPointer value: %p, points at value: %d\n\n", (void*)p_cost, *p_cost);

---

No more warnings given by the compiler. Thanks a lot :-))

Quick follow up question about the output of the program, if I may.

Output of printf("\nPointer value: %d, points at value: %d\n\n", p_cost, *p_cost);

Pointer value: 2293620, points at value: 100

Output of printf("\nPointer value: %p, points at value: %d\n\n", (void*)p_cost, *p_cost);

Pointer value: 0022FF74, points at value: 100

Why is the output different? Is it just a different way of indicating a pointer value?

Thanks.

The first one shows the pointer as a decimal number, the second shows it in hexadecimal notation. It's the same value (I think - it's about 2.2 million from the hex number - since you have recompiled, it's not guaranteed to be the same address anyways).

It is actually not strictly defined what %p outputs, just that it's compatible with displaying pointers. To most people, the value itself is not very important, it's more a case of understanding which pointer is which or the difference between pointers.

--
Mats

Thanks a lot. It's clear as day now :-))