默认函数参数可以“填充”扩展参数包吗？

Question

The following code fails to compile :

#include <iostream>

template<typename F, typename ...Args>
static auto wrap(F func, Args&&... args)
{
    return func(std::forward<Args>(args)...);
}

void f1(int, char, double)
{
    std::cout << "do nada1\n"; 
}

void f2(int, char='a', double=0.)
{
    std::cout << "do nada2\n"; 
}

int main()
{
    wrap(f1, 1, 'a', 2.); 
    wrap(f2, 1, 'a'); 
}

g++ -std=c++14 -O2 -Wall -pedantic -pthread main.cpp && ./a.out

main.cpp: In instantiation of 'auto wrap(F, Args&& ...) [with F = void(*)(int, char, double); Args = {int, char}]':
main.cpp:22:20:   required from here
main.cpp:6:44: error: too few arguments to function
     return func(std::forward<Args>(args)...);

It seems that the rule about "parameter packs being last" is followed (at least in the declaration) and after the expansion a correct function call should be formed : f2 can either be called with 1, 2 or 3 arguments so the too few arguments being an error seems 'harsh'. It also doesn't look like a deduction problem (which would be my guess - but got shaky due to the error message)

Is this a missing feature or there's a violation from the Standard's point of view ?

Answer 1

You aren't calling a function with default-arguments from the template.

You are calling a function-pointer, which points to a function expecting exactly 3 arguments, neither more nor less.

Of course the compiler complains bitterly about the missing third one.

You could do what you are trying to do there with a variadic functor, since C++14 even a lambda :

wrap([](auto&&... args){return f2(std::forward<decltype(args)>(args)...);}, 1, 'a');