Why is "operator void" not invoked with cast syntax?
c++, casting, operator-overloading, operators
Solution
The technical reason why is found in §12.3.2:
A conversion function is never used to convert a (possibly cv-qualified) object to the (possibly cv-qualified) same object type (or a reference to it), to a (possibly cv-qualified) base class of that type (or a reference to it), or to (possibly cv-qualified) void.
The rationale is (likely) to allow §5.2.9/4 to work:
Any expression can be explicitly converted to type “cv void.” The expression value is discarded.
`(void)expr` to suppose to do nothing for the resulting value of any expression, but if it called your conversion operator it wouldn't be discarding anything. So they ban the use of `operator void` in conversions.
Why not make it ill-formed to have the conversion-type-id be `void`? Who knows, but keep in mind it's not totally useless:
struct foo
{
operator void()
{
std::cout << "huh?" << std::endl;
}
};
typedef void (foo::*void_function)();
foo f;
void_function func = &foo::operator void;
(f.*func)(); // prints "huh"
f.operator void(); // also does (which you knew)
It is still technically potentially useful for something, so maybe that's rationale enough not to make it ill-formed.
Problem
While playing with this answer by user GMan I crafted the following snippet (compiled with Visual C++ 9): ``` class Class { public: operator void() {} }; Class object; static_cast<void>( object ); (void)object; object.operator void(); ``` after stepping over with the debugger I found out that casting to `void` doesn't invoke `Class::operator void()`, only the third invokation (with explicitly invoking the operator) actually invokes the operator, the two casts just do nothing. Why is the `operator void` not invoked with the cast syntax?