How delegate from copy constructor to universal copy constructor template?

问题

If I want to write a universal copy constructor (one that will take any argument type), it's easy enough to do:

class Widget {
public:
  template<typename T> Widget(T&& other);
};

This won't prevent the compiler from generating a traditional copy constructor, so I'd like to write that myself and delegate to the template. But how do I do that?

class Widget {
public:
  template<typename T> Widget(T&& other);
  Widget(const Widget& other): ??? {}         // how delegate to the template?
};

I tried to write the delegating constructor this way,

Widget(const Widget& other): Widget<const Widget&>(other){}

but VC11, gcc 4.8, and clang 3.2 all reject it.

How can I write the delegating copy constructor I'm trying to write?

回答1:

You can't specify template arguments for a constructor template; you're limited to deduction. Maybe you can adapt an answer from one of my old posts:

// NOT Tested!
#include <utility>

class Widget
{
    enum fwd_t { fwd };

    // Your true master constructor
    template< typename T >  Widget( fwd_t, T &&t );

public:
    template < typename T >
    Widget( T&& other );
        : Widget( fwd, std::forward<T>(other) )
    {}
    Widget( const Widget& other )
        : Widget( fwd, other )
    {}
};

The two single-argument constructors forward to a two-argument sibling constructor.

来源：https://stackoverflow.com/questions/15264252/how-delegate-from-copy-constructor-to-universal-copy-constructor-template