Thread: sizeof unsigned long long int

I am trying to figure out how many bytes are in an unsigned long long int on my desktop computer. But I keep getting warnings:

warning: format ‘%d’ expects type ‘int’, but argument 2 has type ‘long unsigned int’

Code:

#include <stdio.h>
int main(void)
{
	unsigned char unit_id[] = { 0x2B, 0xC, 0x6B, 0x54}; // 8-bit
	unsigned long long int unit_id_val;
	int i;
    printf("the size of unsigned long long int %d", sizeof(unit_id_val  ) );


	return 0;
}

Any idea how to fix the warnings?

sizeof returns a type size_t:

Originally Posted by C99 6.5.3.4 p4

4 The value of the result is implementation-defined, and its type (an unsigned integer type)
is size_t, defined in <stddef.h> (and other headers).

size_t is not necessarily an int, so %d is not necessarily the right format. Thankfully, we can look up the right one:

Originally Posted by C99 7.19.6.1 p7

z Specifies that a following d, i, o, u, x, or X conversion specifier applies to a
size_t or the corresponding signed integer type argument; or that a
following n conversion specifier applies to a pointer to a signed integer type
corresponding to size_t argument.

So, in order to print a size_t, you need to use the z length modifier, something like:

Code:

printf("the size of unsigned long long int %zd", sizeof(unit_id_val));

EDIT: Dangit! You're too fast laserlight, that's the second time you ninja-posted me today.