C output question

Question

Why its output is %%??

#include
int main(void)
{
        printf(\"% % %\\n\");
return 0;
}

Answer 1

It's undefined behaviour and absolutely anything can happen. Section 7.19.6.1/9 of C99 states:

If a conversion specification is invalid, the behavior is undefined.

and none of the preceding sections allow a conversion specifier of a space. They are limited to characters from the set diouxXfFeEgGaAcsPn%.

Answer 2

If you use one %, it sees it as string (because it lacks other specifiers) and output %. If you use %%, it is to print % in output. if you use %%% the first two will be considered as outputting % and the last one as single "character". so you only get two %.