Mmh, who are you PRIu64?

Question

I am new to C and I am confronted with:

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

int main(void)
{
    uint64_t foo = 10;
    printf("foo is equal to %" PRIu64 "!\n", foo);
    
    return 0;
}

And it works! I don't understand why. Can somebody help me about this?

Answer 1

PRIu64 is a format specifier, introduced in C99, for printing uint64_t, where uint64_t is (from linked reference page):

unsigned integer type with width of ... 64 bits respectively (provided only if the implementation directly supports the type)

PRIu64 is a string (literal), for example the following:

printf("%s\n", PRIu64);

prints llu on my machine. Adjacent string literals are concatenated, from section 6.4.5 String literals of the C99 standard:

In translation phase 6, the multibyte character sequences specified by any sequence of adjacent character and wide string literal tokens are concatenated into a single multibyte character sequence. If any of the tokens are wide string literal tokens, the resulting multibyte character sequence is treated as a wide string literal; otherwise, it is treated as a character string literal.

This means:

printf("foo is equal to %" PRIu64 "!\n", foo);

(on my machine) is the same as:

printf("foo is equal to %llu!\n", foo);

See http://ideone.com/jFvKR9 .