Yes, it is valid.
As for making it work with functors as well, the usual solution is something like this instead:
template <typename F>
void doOperation(F f)
{
int temp=0;
f(temp);
std::cout << "Result is " << temp << std::endl;
}
which can now be called as either:
doOperation(add2);
doOperation(add3());
The problem with this is that if it makes it tricky for the compiler to inline the call to add2
, since all the compiler knows is that a function pointer type void (*)(int &)
is being passed to doOperation
. (But add3
, being a functor, can be inlined easily. Here, the compiler knows that an object of type add3
is passed to the function, which means that the function to call is add3::operator()
, and not just some unknown function pointer.)