Why function default parameters cannot be perferct forwarded in C++?

Question

I found quite weird behavior from my perspective: function default arguments cannot be forwarded in code bellow.

void Test(int test = int{}) {}

template<typename F, typename ...Args>
void Foo(F&& f, Args&&... args)
{
    std::forward<F>(f)(std::forward<Args>(args)...);
}

int main()
{
    Foo(Test, 0); // This compiles
    Foo(Test);    // This doesn't compile
}

Clang reports: error: too few arguments to function call, expected 1, have 0 GCC and VC report same errors.

Can anybody explain it?

Code is here: http://rextester.com/live/JOCY22484

Answer 1

Test is a function that always takes one argument. If its declaration with the default argument is visible when Test is invoked by name, the compiler will implicitly add the default argument to the call. However, once Test has been converted into a pointer or reference to function, the default argument information is no longer visible.

This can be worked around by creating a functor which really does take zero or one arguments and has that information encoded into its type so it cannot be destroyed, like so:

struct Test {
    void operator()(int) { /* ... */ }
    void operator()() { operator(int{}); }
} test;
// ...
Foo(test, 0); // ok
Foo(test);    // ok