Why cast unused return values to void?

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 c

Answer 1

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(fn());

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