在这个例子中移动和前进的区别

Question

The first example that take A by value does two moves and the one by refref only does one move. What is the difference?

struct A
{
  A() { cout << "constructor" << endl;}
  A(const A&) { cout << "copy constructor " << endl;}
  void operator=(const A&) { cout << "assignment operator" << endl; }
  A( A&&) { cout << "move copy constructor" << endl;}
  void operator=(A&&) { cout << "move assignment operator" << endl;}
};
struct C {
  void func(A t) {
    d.a = std::move(t);
  }
  struct Data {
    A a;      
  };
  Data d;
};
struct B {
  void func(A t) {
    C c;
    c.func(std::move(t));
  }
};
//////////////////////////////////////////////////////////
struct C {
  template<class T>
  void func(T&& t) {
    d.a = std::forward<T>(t);
  }
  struct Data {
    A a;      
  };
  Data d;
};
struct B {
  template<class T>
  void func(T&& t) {
    C c;
    c.func(std::forward<T>(t));
  }
};

Answer 1

From cppreference.com :

When used according to the following recipe in a function template, forwards the argument to another function exactly as it was passed to the calling function.

 template<typename T> wrapper(T&& arg) { foo(std::forward<T>(arg)); }

So in your snippet

struct B {
  template<class T>
  void func(T&& t) {
    C c;
    c.func(std::forward<T>(t));
  }
};

The std::foward<T>(t) will simply forward your T&& object to c.func() exactly as B::func() was called. This doesn't require a move, which is why you are seeing fewer moves using std::forward<T> .

I would really recommend checking out Scott Meyer's blog post on this topic of std::move and std::forward : http://scottmeyers.blogspot.com/2012/11/on-superfluousness-of-stdmove.html