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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 ); } 
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So yeah, both are same; the only difference is that C++ defines size_t in std namespace.
std::size_t is the unsigned integer type of the result of the sizeof operator as well as the sizeof... operator and the alignof operator (since C++11). The bit width of std::size_t is not less than 16. (since C++11)
size_t is a base unsigned integer memsize-type defined in the standard library of C/C++ languages. This type is described in the header file stddef.
The name size_t essentially means "size type", and you will typically see this data type used to specify the size or length of things - like the length of a C string returned by the strlen() function, for example. This is not one of the "built-in" data types of C/C++.
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.
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