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

c++ size-t
jwfearn picture jwfearn · Oct 26, 2008 · Viewed 31.9k times · Source

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

Johannes Schaub - litb picture Johannes Schaub - litb · Nov 12, 2008

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.

Related questions

unsigned int vs. size_t

I notice that modern C and C++ code seems to use size_t instead of int/unsigned int pretty much everywhere - from parameters for C string functions to the STL. I am curious as to the reason for this …

Read More
How to cast the size_t to double or int C++

My question is that I have a size_t data, but now I want to convert it to double or int. If I do something like size_t data = 99999999; int convertdata = data; the compiler will report warning. because it maybe …

Read More
should use size_t or ssize_t

At my code, I do not use int or unsigned int. I only use size_t or ssize_t for portable. For example: typedef size_t intc; // (instead of unsigned int) typedef ssize_t uintc; // (instead of int) Because strlen, …

Read More