Question

Possible Duplicate:
Why cast unused return values to void?

What is the purpose of (void) before a function call, for example

(void) func1();

I assume this is the same as simply calling func1();

Therefore is the (void) call simply to let other programmers know that the return type will be ignored, for instance if func1() had a return type of int, or does the compiler perhaps perform some optimizations on the function? Perhaps there is another reason behind it altogether – is it even legal C++ or perhaps is it a remnant of C seen in some legacy code.

Thanks

Answer 1

A cast to void can have semantic effect in one case: where a value is an operand of the comma operator and overrides the comma operator, a cast to void will suppress it:

struct S {
  int operator,(int) { return 0; }
};
std::cout << (S(), 42) << '\n';           // prints '0'
std::cout << ((void) S(), 42) << '\n';    // prints '42'