Page 169 - C-Language

Chapter 23: Formatted Input/Output

Examples

Printing the Value of a Pointer to an Object

To print the value of a pointer to an object (as opposed to a function pointer) use the p conversion
specifier. It is defined to print void-pointers only, so to print out the value of a non void-pointer it
needs to be explicitly converted ("casted*") to void*.

#include <stdlib.h> /* for EXIT_SUCCESS */
#include <stdio.h> /* for printf() */

int main(void)
{
int i;
int * p = &i;

printf("The address of i is %p.\n", (void*) p);

return EXIT_SUCCESS;
}

C99

Using <inttypes.h> and uintptr_t

Another way to print pointers in C99 or later uses the uintptr_t type and the macros from
<inttypes.h>:

#include <inttypes.h> /* for uintptr_t and PRIXPTR */
#include <stdio.h> /* for printf() */

int main(void)
{
int i;
int *p = &i;

printf("The address of i is 0x%" PRIXPTR ".\n", (uintptr_t)p);

return 0;
}

In theory, there might not be an integer type that can hold any pointer converted to an integer (so
the type uintptr_t might not exist). In practice, it does exist. Pointers to functions need not be
convertible to the uintptr_t type — though again they most often are convertible.

If the uintptr_t type exists, so does the intptr_t type. It is not clear why you'd ever want to treat
addresses as signed integers, though.

K&RC89

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