Can't understand move semantics

Question

In a simple example like the following:

struct MyClass {
    MyClass() {}
    MyClass(const MyClass&) {}
};

int main(){
    std::vector<MyClass> objects;
    MyClass obj1;
    objects.push_back(obj1);
    for (auto&& obj: object) {
    }
    return 0;
}

how does the auto&& work? The class doesn't have any move support, what is this code supposed to do? I don't understand..

possible duplicate of [What is the advantage of using universal references in range-based for loops?](http://stackoverflow.com/questions/13130708/what-is-the-advantage-of-using-universal-references-in-range-based-for-loops) — The Paramagnetic Croissant, Oct 27 '14 at 16:00
@juanchopanza fine, but why does it compile? What is auto&& in that case? — Dean, Oct 27 '14 at 16:03

score 1 · Accepted Answer · answered Oct 27 '14 at 16:08

The type is deduced from the type obtained by dereferencing the sequence's iterator type; that is an lvalue reference, MyClass&. By the "reference collapsing" rules, trying to deduce the type "rvalue reference to lvalue reference" gives the lvalue reference type.

So auto&& is deduced to be MyClass&.

If the sequence had an iterator type that gave rvalue references, then it would instead be deduced as MyClass&&, enabling move semantics if the class supports them.

Can't understand move semantics

1 Answers1