Lvalue reference constructor is called instead of rvalue reference constructor

问题

There is this code:

#include <iostream>

class F {
public:
   F() = default;
   F(F&&) {
      std::cout << "F(F&&)" << std::endl;
   }
   F(F&) {
      std::cout << "F(F&)" << std::endl;
   }
};

class G {
   F f_;
public:
   G(F&& f) : f_(f) {
      std::cout << "G()" << std::endl;
   }
};

int main(){
   G g = F();
   return 0;
}

The output is:

F(F&)
G()

Why F(F&) constructor is called instead of F(F&&) constructor in constructor of class G? The parameter for constructor of class G is F&& f which is rvalue reference but constructor for lvalue reference is called.

回答1:

Why F(F&) constructor is called instead of F(F&&) constructor in constructor of class G?

Because f is an lvalue. Even though it is bound to an rvalue, and its type is rvalue reference to F, it is also a named variable. That makes it an lvalue. Don't forget that the value category of an object is not determined by its type, and vice versa.

When you pass an lvalue to a function, only lvalue references can be bound to it. You should change your code as follows if you want to catch rvalues only:

class G {
    F f_;
public:
    G(F&& f) : f_(std::move(f)) {
       std::cout << "G()" << std::endl;
    }
};

Alternatively, you could use std::forward<>(), which is equivalent in this case, but makes your intent of forwarding f even clearer:

class G {
    F f_;
public:
    G(F&& f) : f_(std::forward<F>(f)) {
       std::cout << "G()" << std::endl;
    }
};

Now this last definition is easy to extend so that both lvalues and rvalues of type F can be bound to the parameter f:

class G {
    F f_;
public:
    template<typename F>
    G(F&& f) : f_(std::forward<F>(f)) {
       std::cout << "G()" << std::endl;
    }
};

This allows, for instance, to construct an instance of G this way:

F f;
G g(f); // Would not be possible with a constructor accepting only rvalues

This last version has a caveat though: your constructor will basically work as a copy-constructor as well, so you might want to explicitly define all the possible copy constructors to avoid awkward situations:

class G {
    F f_;
public:
    template<typename F>
    G(F&& f) : f_(std::forward<F>(f)) {
       std::cout << "G()" << std::endl;
    }
    G(G const&) = default;
    G(G&); // Must be defaulted out-of-class because of the reference to non-const
};

G::G(G&) = default;

Since non-template functions are preferred over instantiations of function templates, the copy constructor will be selected when constructing a G object from another G object. The same applies, of course, to the move constructor. This is left as an exercise.

来源：https://stackoverflow.com/questions/15314727/lvalue-reference-constructor-is-called-instead-of-rvalue-reference-constructor

标签

c++

c++11

move-semantics

rvalue-reference