tipo de retorno de std::bind implícitamente convertible a dos constructores explícitos diferentes

Dado dos explicit constructor overloads (based on different std::function<...> types), the return value of std::bind es capaz de seleccionar ya sea (thereby making the call ambiguous)

call of overloaded ‘Bar(std::_Bind_helper<false, void (Foo::*)(int), 
    Foo*, int>::type)’ is ambiguous

If I comment out either, then the code compiles!

I would have thought making the constructors explicit would have either selected the correct overload, or prevented both from being selected?

Por supuesto explícitamente creando un std::function at the point I bind works:

    Bar b(std::function<void(int)>(std::bind((&Foo::process), &f, 1)));

However, I'm puzzled as to why type deduction doesn't work?

• If the return value from std::bind matches neither of the two constructor signatures, the fact they are explicit should prevent both from being selected.
• If the return value from std::bind matches one of the two constructor signatures, the fact they are explicit should cause the correct one to be selected.

¿Qué está pasando realmente aquí?

Full working code below:

#include <functional>

struct Foo
{
    void process(int) { }
};

struct Bar
{
    // comment out either of these to compile
    explicit Bar(std::function<void(int)>) {} 
    explicit Bar(std::function<void(short)>) {}
};

int main()
{
    Foo f;
    Bar b(std::bind(&Foo::process, &f, 1));
    return 0;
}