Question

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?

Answer 1

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.