Is the %zu specifier required for printf?

Question

We are using C89 on an embedded platform. I attempted to print out a size_t, but it did not work:

#include <stdio.h>
int main(void) {
    size_t n = 123;
    printf("%zu\n",n);
    return 0;
}

Instead of 123, I got zu.
Other specifiers work correctly.

If size_t exists shouldn't zu also be available in printf?
Is this something I should contact my library vendor about, or is a library implementation allowed to exclude it?

Answer 1

If size_t exists shouldn't zu also be available in printf?

size_t existed at least since C89 but the respective format specifier %zu (specifically the length modifier z) was added to the standard only since C99.

So, if you can't use C99 (or C11) and had to print size_t in C89, you just have to fallback to other existing types, such as:

printf("%lu\n", (unsigned long)n);

Answer 2

size_t is returned by sizeof(). %zu specifier prints the length.So to return the size of n,the code should read: #include <stdio.h>

int main(void) { size_t n=123; printf("%zu\n",sizeof (n)); return 0;` }

Answer 3

This is working fine. In my system, it's printing the expected value. You may be getting the error due to the compiler. I am using the GCC compiler. If you have this only and trying to get the output the first update the GCC and then try. It will work.