为什么（无效）0在C和C ++无操作？ [英] Why is (void) 0 a no operation in C and C++?

问题描述

我见过的的glibc 的调试用printfs在内部被定义为（无效）0 ，如果 NDEBUG 的是定义。同样， __空操作 为Visual C ++编译器也在那儿。在GCC和VC ++编译器，前者的作品，而后者只能在VC ++。现在大家都知道，无论是上述声明将被视为无操作，也没有相应的code将产生;但这里的地方我有一个疑问。

I have seen debug printfs in glibc which internally is defined as (void) 0, if NDEBUG is defined. Likewise the __noop for Visual C++ compiler is there too. The former works on both GCC and VC++ compilers, while the latter only on VC++. Now we all know that both the above statements will be treated as no operation and no respective code will be generated; but here's where I've a doubt.

在的情况下，__空操作，MSDN说，这是由编译器提供的内部函数。来到（无效）0 〜为什么间$ P $由编译器为无操作PTED？它是C语言的一个棘手的用法还是标准说的一些东西明确地？甚至可以说，是值得做的编译器实现？

In case of __noop, MSDN says that it's a intrinsic function provided by the compiler. Coming to (void) 0 ~ Why is it interpreted by the compilers as no op? Is it a tricky usage of the C language or does the standard say something about it explicity? Or even that is something to do with the compiler implementation?

推荐答案

（无效）0 （+ ; ）是合法的，但不会无的C ++前pression，这就是一切。它不翻译成目标架构的无操作指令，它只是一个空的语句作为占位符每当语言需要一个完整的语句（例如目标为一跳转标签，或在如果条款的机构）。

(void)0 (+;) is a valid, but 'does-nothing' C++ expression, that's everything. It doesn't translate to the no-op instruction of the target architecture, it's just an empty statement as placeholder whenever the language expects a complete statement (for example as target for a jump label, or in the body of an if clause).

编辑：（根据克里斯·卢茨的评论更新）

(updated based on Chris Lutz's comment)

应当注意的是，作为宏使用时，说

It should be noted that when used as a macro, say

#define noop ((void)0)

在（无效） $ P $被意外使用像

the (void) prevents it from being accidentally used as a value like

int x = noop;

有关上述前pression编译器会正确地将其标记为无效操作。 GCC吐​​奶错误：空值不被忽略，因为它应该是和VC ++树皮无效非法拥有各类。

For the above expression the compiler will rightly flag it as an invalid operation. GCC spits error: void value not ignored as it ought to be and VC++ barks 'void' illegal with all types.

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