为什么将未使用的返回值强制转换为 void?

int fn();

void whatever()
{
    (void) fn();
}

Is there any reason for casting an unused return value to void, or am I right in thinking it's a complete waste of time?

Follow up:

Well that seems pretty comprehensive. I suppose it's better than commenting an unused return value since self documenting code is better than comments. Personally, I'll turn these warnings off since it's unnecessary noise.

I'll eat my words if a bug escapes because of it...

解决方案

David's answer pretty much covers the motivation for this, to explicitly show other "developers" that you know this function returns but you're explicitly ignoring it.

This is a way to ensure that where necessary error codes are always handled.

I think for C++ this is probably the only place that I prefer to use C-style casts too, since using the full static cast notation just feels like overkill here. Finally, if you're reviewing a coding standard or writing one, then it's also a good idea to explicitly state that calls to overloaded operators (not using function call notation) should be exempt from this too:

class A {};
A operator+(A const &, A const &);

int main () {
  A a;
  a + a;                 // Not a problem
  (void)operator+(a,a);  // Using function call notation - so add the cast.