来自< stddef.h>的'offsetof'宏调用未定义的行为？ [英] Does the 'offsetof' macro from <stddef.h> invoke undefined behaviour?

问题描述

MSVC实现的示例：

Example from MSVC's implementation:

#define offsetof(s,m) \
    (size_t)&reinterpret_cast<const volatile char&>((((s *)0)->m))
//                                                   ^^^^^^^^^^^

可以看出，它解引用一个空指针，通常调用未定义的行为。

As can be seen, it dereferences a null pointer, which normally invokes undefined behaviour. Is this an exception to the rule or what is going on?

推荐答案

语言标准说未定义的行为时，任何给定的编译器可以定义行为。标准库中的实现代码通常依赖于它。所以有两个问题：

Where the language standard says "undefined behavior", any given compiler can define the behavior. Implementation code in the standard library typically relies on that. So there are two questions:

（1）代码UB是否与C ++标准相关？

(1) Is the code UB with respect to the C++ standard?

这是一个非常困难的问题，因为它是一个众所周知的几乎缺陷，C ++ 98/03标准从来没有说在规范文本，一般来说，它是UB解引用空指针。这是 由 typeid 的例外，其中不是 UB。

That's a really hard question, because it's a well known almost-defect that the C++98/03 standard never says right out in normative text that in general it's UB to dereference a nullpointer. It is implied by the exception for typeid, where it's not UB.

你可以肯定地说，它是UB使用非POD类型 offsetof 。

What you can say decidedly is that it's UB to use offsetof with a non-POD type.

（2）是关于它编写的编译器的代码UB？

(2) Is the code UB with respect to the compiler that it's written for?

否，当然不是。

编译器供应商给定编译器的代码可以使用该编译器的任何特性。

A compiler vendor's code for a given compiler can use any feature of that compiler.

hth。，

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