int类型的sizeof是如何工作的？ [英] How does sizeof work for int types?

问题描述

我有一个小程序比较

（1）sizeof，
（2）numeric_limits :: digits，
（3）一个循环的结果

(1) sizeof, (2) numeric_limits::digits, (3) and the results of a loop

，以确保他们都报告关于任何C ++实现的int类型的大小相同的事情。但是因为我不知道sizeof的内部，我不知道是否只是报告numeric_limits ::位数。感谢

in an effort to make sure they all report the same thing regarding the size of the "int types" on any C++ implementation. However because I don't know about the internals of sizeof, I have to wonder if it is just reporting numeric_limits::digits. Thanks

推荐答案

在大多数编译器上最可能的 sizeof（）在其内部类型表中查找给定类型（或对象类型），并将该类型的定义大小的文本插入到它生成的代码中。 这将发生在编译时，而不是运行时。

Most likely sizeof() on most compilers causes the compiler to look the given type (or object's type) up in its internal type table and insert a literal for that type's defined size into the code it generates. This would happen at compile time, not runtime.

要回答注释中的问题，没有任何语言定义的访问编译器的内部在C ++中（当然，除了 sizeof（）本身）。我知道的唯一类似的语言就是Ada，它提供了 ASIS 用于编写与编译器无关的代码分析工具。

To answer the question in the comments, there isn't any language-defined access to the compiler's internals in C++ (outside of things like sizeof() itself, of course). The only similar language I know of that lets you do stuff like that is Ada, which provides ASIS for writing compiler-independent code analysis tools.

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