为什么 (void) 0 在 C 和 C++ 中是无操作

发布于02月04日

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.

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

(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).

EDIT: (updated based on Chris Lutz's comment)

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

#define noop ((void)0)

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

int x = noop;

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.

为什么 (void) 0 在 C 和 C++ 中是无操作

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