CODEWITHSUNDEEP
c++c++11c++14
edaniels
edaniels

Reputation: 778

What does it mean to take the address of an rvalue reference?

Given

void foo( int&& x ) {

    std::cout << &x;
}

This works but what does this address actually represent? Is a temporary int created when foo is called and that is what the address represents? If this is true and if I write int y = 5; foo(static_cast<int&&>(y));, does this cause another temporary to be created or will the compiler intelligently refer to y?

Upvotes: 19

Views: 4337

Answers (1)

Anton Savin
Anton Savin

Reputation: 41321

When you take an address of an rvalue reference, it returns a pointer to the object that reference is bound to, like with lvalue references. The object can be a temporary or not (if for example you cast an lvalue to rvalue reference like you do in your code).

int y = 5; foo(static_cast<int&&>(y)); does not create a temporary. This conversion is described in part 5.2.9/3 of the standard:

A glvalue, class prvalue, or array prvalue of type “cv1 T1” can be cast to type “rvalue reference to cv2 T2” if “cv2 T2” is reference-compatible with “cv1 T1”.

A temporary will be created if for example you call foo(1);.

Upvotes: 14

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