[cfe-dev] Move constructor forces copy assignment to be implicitly defaulted?

Steve Ramsey clang at lucena.com
Mon May 28 16:04:52 PDT 2012

On May 29, 2012, at 1:56 AM, Suman Kar (from the future) wrote:

> A quick question:
> struct A {};
> struct B : A { ~B() {} };
> A makeA() {
> A a;
> return a;
> }
> B makeB() {
> B b;
> return b;
> }
> int main() {
> A a = makeA();
> B b = makeB();
> }
> Is it correct to assume that A will be moved and B will not? Should a
> diagnostic be required for the call to makeB?

I’m real glad you asked that. This plays into 12.8 p7 from the FDIS:

> If the class definition does not explicitly declare a copy constructor, one is declared implicitly. If the class definition declares a move constructor or move assignment operator, the implicitly declared copy constructor is defined as deleted; otherwise, it is defined as defaulted (8.4). The latter case is deprecated if the class has a user-declared copy assignment operator or a user-declared destructor.

This paragraph features in my nightmares. I go back every few weeks and try and parse it again, but I still get confused. In my current interpretation, it’s saying that a copy constructor will always be declared, whether implicitly or explicitly; any implicitly declared copy constructor will always be defined as deleted unless there are no implicitly or explicitly declared move special member functions.

“But, wait,” you’re thinking, “didn’t I ask about the move constructor?” Well, yes. B’s move constructor will not be declared at all because of the presence of a user-declared destructor, and hence will not be defined. However, B is still technically move-constructible because it will fall back to the implicitly defined copy constructor (as per the previously mentioned Note in 12.8 p9), and that copy constructor exists because of 12.8 p7.


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