Output address of variable without printf %p

Question

I am a first year student in C programming so my skill/knowledge is limited. I am trying to create my own implementation on printf, but I am having trouble with retrieving and printing the address of a variable. With printf, its possible to output the address of a variable with %p, i need to replicate %p somehow.

When storing the address of a variable, the data type is int*, and I cannot figure out how to write this address to the screen(stdout).

For Example:

int i        = 123;
int *address = &i;

Now how would I output address(not the value at i)? I have tried using the original printf format specifiers for testing purposes. I tried using %x, %s, %d, %lu... it all gives me an error as I am trying to output an int*(integer pointer).

Can anyone assist me in outputting the address?

@Anders K. This will cut the upper 32bit of a 64bit address on a 64bit machine. — kwarnke, May 27 '16 at 10:52

2501 · Answer 1 · 2016-05-27T10:56:09.520

5

You can only inspect the bits of the pointer and print those, using the type unsigned char:

unsigned char* a = ( unsigned char* )address;
for( size_t i = 0 ; i < sizeof( address ) ; i++ )
     printf( "%hhu" , a[i] );

Another option, if the pointer is a pointer to an object, is to cast the pointer to types: intptr_t or uintptr_t, but the availability of those types is implementation defined.

edited May 27 '16 at 10:56

answered May 27 '16 at 10:34

2501

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They should be available since C99. – Aconcagua May 27 '16 at 10:41
1

@Aconcagua Those types are optional, thus it depends on the implementation. – 2501 May 27 '16 at 10:42

score 4 · Accepted Answer · edited May 23 '17 at 12:15

4

If it's for the address of an object you could do the following:

#include <stdio.h>
#include <stdint.h>
#include <inttypes.h>

int main(void)
{
  int a;
  uintptr_t uip = (uintptr_t) ((void*) &a);
  printf("address of a = 0x%"PRIXPTR, uip);
}

To print the address of a function you need to go as proposed here.

edited May 23 '17 at 12:15

Community

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answered May 27 '16 at 10:41

alk

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I didn't find anything that would prohibit casting function pointers to uintptr_t. Did I miss something? – 2501 May 27 '16 at 10:45
1

I already did: *Any pointer type may be converted to an integer type.* Since a distinction is made in other rules between object and function pointer types, I think *any pointer type* applies to both. – 2501 May 27 '16 at 10:47
1

@2502: Ok, got it: `intptr_t` and `uintptr_t` are guaranteed by the Standard only to hold the value of a valid `void*` (7.20.1.4/1). A function may not be converted to a `void*`. – alk May 27 '16 at 10:52
@2501: Sry, for misspelling your name. Interesting typo however ... ;-) – alk May 27 '16 at 10:59
So I deduce: Casting to void* previously is obsolete and you simply could have `(uintptr_t) &a;`... – Aconcagua May 27 '16 at 11:01
@Aconcagua: "*Casting to void\* previously is obsolete*" it isn't. Please see my comment above quoting the C Standard. – alk May 27 '16 at 11:22
@alk I deduced from 2501 citation: *"Any pointer type may be [...]"* - the void* part in your comment is in my eyes rather a size guarantee: It can hold a valid void*, thus can hold any other valid pointer, too, as long as it is convertible to void* (and thus excludes function pointers). – Aconcagua May 27 '16 at 11:37

Output address of variable without printf %p

For Example:

2 Answers2