C – Is the %zu Specifier Required for printf?

c-librariesc++c89printfsize-t

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?

Best Answer

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);

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C printf – Correct Way to Use printf to Print a size_t

Try using the %zu format string

size_t val = get_the_value();
printf("%zu",val);

The z portion is a length specifier which says the argument will be size_t in length.

Source - http://en.wikipedia.org/wiki/Printf#printf_format_placeholders

C++ Printf Format Specifier for size_t: %zu or %Iu?

MS Visual Studio didn't support %zu printf specifier before VS2013. Starting from VS2015 (e.g. _MSC_VER >= 1900) %zu is available.

As an alternative, for previous versions of Visual Studio if you are printing small values (like number of elements from std containers) you can simply cast to an int and use %d:

printf("count: %d\n", (int)str.size()); // less digital ink spent
// or:
printf("count: %u\n", (unsigned)str.size());

Best Answer

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C printf – Correct Way to Use printf to Print a size_t

C++ Printf Format Specifier for size_t: %zu or %Iu?

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