Get an rvalue when calling a getter method on an rvalue object

Question

Suppose, I have the following code. There's a copy constructor in B which calls a method which copies the resources of a.

Now I also have a move constructor. In this case, a should not be copied but just "steal" the resources from an existing a. Therefore, I also implemented an init taking an rvalue. But of course, when I try to call it with parameter b.a, this is an lvalue...

Is there a way to call this method?

class A{

    A(const A&& a){
        // 'steal' resources from a
    }

    void init(A& a){
       // init this A from another A by copying its resources
    }

    void init(A&& a){
      // init this A from another A stealing its resources and tell the other a, it must not destroy resources upon destruction
    }
};

class B{
    A a;

    B(B& b){
      a.init(b.a)          
    }

    B(B&& b){
      a.init(b.a); // How to call init(A&& a)?  
    }

};

juanchopanza · Accepted Answer · 2014-07-23T16:05:27.633

7

b.a is an lvalue, so you need to apply std::move:

a.init(std::move(b.a));

Note: But why is b an lvalue in the body of B(B&& b)?

Here, the parameter type B&& b simply means that this constructor overload will be chosen over, say, B(const B& b) when invoked with an rvalue.

B make_B() { return B(); }
B b1(make_B());            // B(B&&) chosen
B b2(b);                   // B(const B&) chosen

But the parameter itself is an lvalue because it has a name. All std::move does is make its argument look like an rvalue.

edited Jul 23 '14 at 16:05

answered Jul 23 '14 at 15:34

juanchopanza

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How is `b.a` an lvalue? – David G Jul 23 '14 at 15:43
@0x499602D2 `b` is an lvalue, `a` is the name of its data member, so `b.a` is an lvalue. – juanchopanza Jul 23 '14 at 15:45
Okay, so how is `b` an lvalue? – David G Jul 23 '14 at 15:46
@0x499602D2 It has a name. – juanchopanza Jul 23 '14 at 15:47
Thanks! You should put that in your answer as the OP most likely made the assumption the `b` is an rvalue too. – David G Jul 23 '14 at 15:49
Yes, I'm confused now.. b in B(B&& b) is an rvalue I think? And: Are names of data members (rvalue.member) of rvalues also rvalues? – Michael Jul 23 '14 at 15:54
1

@Michael `b` is a reference to an rvalue but since it has a name it is treated as an lvalue. `std::move()` is a way to cast that object back into an rvalue (technically xvalue). – David G Jul 23 '14 at 16:03
2

@0x499602D2 I added a short explanation. – juanchopanza Jul 23 '14 at 16:05

Get an rvalue when calling a getter method on an rvalue object

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