Forwarding all constructors in C++0x

问题

What is the correct way to forward all of the parent's constructors in C++0x?

I have been doing this:

class X: public Super {
    template<typename... Args>
        X(Args&&... args): Super(args...) {}
};

回答1:

There is a better way in C++0x for this

class X: public Super {
  using Super::Super;
};

If you declare a perfect-forwarding template, your type will behave badly in overload resolution. Imagine your base class is convertible from int and there exist two functions to print out classes

class Base {
public:
  Base(int n);
};

class Specific: public Base {
public:
  template<typename... Args>
    Specific(Args&&... args);
};

void printOut(Specific const& b);
void printOut(std::string const& s);

You call it with

printOut("hello");

What will be called? It's ambiguous, because Specific can convert any argument, including character arrays. It does so without regard of existing base class constructors. Inheriting constructors with using declarations only declare the constructors that are needed to make this work.

回答2:

I think you need to do Super(std::forward<Args>(args)...) if you want things forwarded properly. args has a name, so I believe it will bind to regular references before rvalue references.

More importantly, though, you won't be forwarding copy constructors this way. The compiler will still generate one for you, because it doesn't consider template constructors. In the example provided, you're okay, because the compiler generated one will just call the parent copy constructor anyhow, which is exactly what you want. If that's not appropriate in a more complicated case, then you'll have to write it yourself.

来源：https://stackoverflow.com/questions/3119929/forwarding-all-constructors-in-c0x