C Language Formatted Input/Output Printing the Value of a Pointer to an Object
Example
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
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
Pre-Standard History:
Prior to C89 during K&R-C times there was no type void* (nor header <stdlib.h>, nor prototypes, and hence no int main(void) notation), so the pointer was cast to long unsigned int and printed using the lx length modifier/conversion specifier.
The example below is just for informational purpose. Nowadays this is invalid code, which very well might provoke the infamous Undefined Behaviour.
#include <stdio.h> /* optional in pre-standard C - for printf() */
int main()
{
int i;
int *p = &i;
printf("The address of i is 0x%lx.\n", (long unsigned) p);
return 0;
}
