Correct printf format specifier for size_t: %zu or %Iu?

Question

I want to print out the value of a size_t variable using printf in C++ using Microsoft Visual Studio 2010 (I want to use printf instead of << in this specific piece of code, so please no answers telling me I should use << instead).

According to the post

Platform independent size_t Format specifiers in c?

the correct platform-independent way is to use %zu, but this does not seem to work in Visual Studio. The Visual Studio documentation at

http://msdn.microsoft.com/en-us/library/vstudio/tcxf1dw6.aspx

tells me that I must use %Iu (using uppercase i, not lowercase l).

Is Microsoft not following the standards here? Or has the standard been changed since C99? Or is the standard different between C and C++ (which would seem very strange to me)?

Answer 1

MS Visual Studio didn't support %zu printf specifier before VS2013. Starting from VS2013 (e.g. _MSC_VER >= 1800) %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());

Answer 2

The Microsoft documentation states:

The hh, j, z, and t length prefixes are not supported.

And therefore %zu is not supported.

It also states that the correct prefix to use for size_t is I – so you'd use %Iu.

Answer 3

Microsoft's C compiler does not catch up with the latest C standards. It's basically a C89 compiler with some cherry-picked features from C99 (e.g. long long). So, there should be no surprise that something isn't supported (%zu appeared in C99).

Answer 4

Based on the answer from here, %z is a C99 addition. Since MSVC doesn't support any of the later C standards, it's no surprise that %z isn't supported.