为什么 (void) 0 在 C 和 C++ 中是无操作?
我在 glibc 中看到了 debug printfs，如果 NDEBUG 被定义，它内部定义为 (void) 0.同样，__noop Visual C++ 编译器也在那里.前者适用于 GCC 和 VC++ 编译器，而后者仅适用于 VC++.现在我们都知道，以上两条语句都将被视为无操作，不会生成相应的代码；但这是我有疑问的地方.
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.
对于__noop，MSDN 说它是编译器提供的内在函数.来到 (void) 0 ~ 为什么它被编译器解释为没有操作?它是 C 语言的一个棘手用法，还是标准明确说明了它?甚至这与编译器实现有关?
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 (+;) 是有效的，但无所事事"的 C++表达，仅此而已.它不会转换为目标体系结构的 no-op 指令，它只是一个空语句作为占位符，只要语言需要一个完整的语句(例如作为跳转标签的目标，或者在if 子句的主体).
(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).
(根据 Chris Lutz 的评论更新)
(updated based on Chris Lutz's comment)
需要注意的是，当用作宏时，说
It should be noted that when used as a macro, say
#define noop ((void)0)
(void) 防止它被意外地用作像
the (void) prevents it from being accidentally used as a value like
int x = noop;
对于上述表达式，编译器会正确地将其标记为无效操作.GCC 吐出 error: void value not ignored as it should be 和 VC++ 吠叫 'void' 对所有类型都是非法的.
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.
相关文章