Why cast unused return values to void?

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.

At work we use that to acknowledge that the function has a return value but the developer has asserted that it is safe to ignore it. Since you tagged the question as C++ you should be using static_cast:

static_cast<void>(fn());

As far as the compiler goes casting the return value to void has little meaning.

The true reason for doing this dates back to a tool used on C code, called lint.

It analyzes code looking for possible problems and issuing warnings and suggestions. If a function returned a value which was then not checked, lint would warn in case this was accidental. To silence lint on this warning, you cast the call to (void).