Mmh, who are you PRIu64?
CMacrosPrintfC Problem Overview
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? Thanks a lot! torr
C Solutions
Solution 1 - C
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 .
Solution 2 - C
You can also use it to print pointer value. Example from LTP BPF patchset:
static uint64_t *val;
...
tst_res(TFAIL, "dst(r7) = %"PRIu64", but should be zero", *val);
prints: dst(r7) = 18446744073709551615, but should be zero
NOTE: tst_res()
is a printf like function.