How does snprintf alignment bytes

Question

Following the answer of here, I wanted to see how many bytes a snprintf uses. The following code

#include "stdio.h"

int main() {
    printf("%d\n", snprintf(NULL, 0, "%d:%llx:%d:%llx:%llx:%llx", 0, 0, 0, 0, 0, 0));
    printf("%d\n", snprintf(NULL, 0, "%d:%llx:%d:%llx:%llx:", 0, 0, 0, 0, 0));
    printf("%d\n", snprintf(NULL, 0, "%llx", 0));
    return 0;
}

returns

22
10
1

I do not understand how the %llx at the end of other data printed can use 12 bytes while it uses only 1 byte if it is used alone. Does snprintf do any byte alignment?

@DavidSchwartz Indeed, never mind, I always forget that C don't infer type in variadic function. — Stargateur, Jun 04 '17 at 06:28

score 3 · Accepted Answer · answered Jun 04 '17 at 06:25

3

Your code has a bug. The %llx format specifier is only for long longs. Try:

#include "stdio.h"

int main() {
    printf("%d\n", snprintf(NULL, 0, "%d:%llx:%d:%llx:%llx:%llx", 0, 0LL, 0LL, 0LL, 0LL, 0LL));
    printf("%d\n", snprintf(NULL, 0, "%d:%llx:%d:%llx:%llx:", 0, 0LL, 0LL, 0LL, 0LL));
    printf("%d\n", snprintf(NULL, 0, "%llx", 0LL));
    return 0;
}

answered Jun 04 '17 at 06:25

David Schwartz

179,497
17
214
278

2

Technically `unsigned long long`, so it should be `0ull`. – melpomene Jun 04 '17 at 06:26
If we do not have ull, how does C compile it? Does it think this is an undefined code? – Joe C Jun 04 '17 at 06:43

How does snprintf alignment bytes

1 Answers1