Does "std::size_t" make sense in C++?

Question

In some code I've inherited, I see frequent use of size_t with the std namespace qualifier. For example:

std::size_t n = sizeof( long ); 

It compiles and runs fine, of course. But it seems like bad practice to me (perhaps carried over from C?).

Isn't it true that size_t is built into C++ and therefore in the global namespace? Is a header file include needed to use size_t in C++?

Another way to ask this question is, would the following program (with no includes) be expected to compile on all C++ compilers?

size_t foo() {     return sizeof( long ); } 

Answer 1

There seems to be confusion among the stackoverflow crowd concerning this

::size_t is defined in the backward compatibility header stddef.h . It's been part of ANSI/ISO C and ISO C++ since their very beginning. Every C++ implementation has to ship with stddef.h (compatibility) and cstddef where only the latter defines std::size_t and not necessarily ::size_t. See Annex D of the C++ Standard.