C：为什么不是size_t型为C的关键字？ [英] C: Why isn't size_t a C keyword?

问题描述

的sizeof 是一个C的关键字的。它返回一个名为为size_t 类型的大小。然而，为size_t 是不会一个关键字，但在主要定义的 STDDEF.H 并可能其他C标准头文件了。

sizeof is a C keyword. It returns the size in a type named size_t. However, size_t is not a keyword, but is defined primarily in stddef.h and probably other C standard header files too.

考虑要建立一个C程序，的场景不会包含任何C标准头文件和库。 （例如像，如果要创建一个操作系统内核）。现在，在这样的code，的sizeof 可以使用（这是一个C的关键字，所以它是的一部分的语言的），但它返回类型（为size_t ）不可用！

Consider a scenario where you want to create a C program which does not include any C standard headers or libraries. (Like for example, if you are creating an OS kernel.) Now, in such code, sizeof can be used (it is a C keyword, so it is a part of the language), but the type that it returns (size_t) is not available!

这是否意味着某种问题，在C标准规范？你能澄清这？

Does not this signify some kind of a problem in the C standard specification? Can you clarify this?

推荐答案

它不从字面上返回size_t类型的值，因为为size_t本身并不是一个具体类型，而是一个typedef到未指定内置类型。 typedef的标识符（如为size_t）是完全等同于它们各自的基础类型（在编译时转换此）。如果size_t会在平台上被定义为一个unsigned int，那么当你的系统上编译的sizeof返回一个unsigned int。为size_t仅仅是保持便携性的一种方便的方法，只需要你明确地使用它的名字被列入STDDEF.H。

It does not literally return a value of type size_t since size_t is not a concrete type in itself, but rather a typedef to an unspecified built-in type. Typedef identifiers (such as size_t) are completely equivalent to their respective underlying types (and are converted thereto at compile time). If size_t is defined as an unsigned int on your platform, then sizeof returns an unsigned int when it is compiled on your system. size_t is just a handy way of maintaining portability and only needs to be included in stddef.h if you are using it explicitly by name.

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