“std :: size_t”在C ++中有意义吗？ [英] Does "std::size_t" make sense in C++?

问题描述

在我继承的一些代码中，我看到 size_t 与 std 命名空间限定词频繁使用。例如：

  std :: size_t n = sizeof（long）; 
  

当然，它编译和运行正常。但是对我来说似乎是不好的做法（可能从C继续。）

不是真的内置在C ++中，因此在全局命名空间？是一个头文件include需要在C ++中使用 size_t ？

另一种提出这个问题的方法是，

$

$

 $ <$ p> 
 c $ c> size_t foo（）
 {
 return sizeof（long）; 
} 
  

解决方案

关于这个 $ :: size_t 的反向兼容性标题 stddef .h 。从它们的开始就是 ANSI / ISO C 和 ISO C ++ 的一部分。每个C ++实现必须附带 stddef.h （兼容性）和 cstddef ，其中只有后者定义 std :: size_t ，而不一定是 :: size_t 。请参阅C ++标准的附录D.

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 );
}

解决方案

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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