"The implicitly-declared or defaulted copy constructor for class T is defined as deleted if any of the following conditions are true ... T has a user-defined move constructor or move assignment operator" (since C++11)

https://en.cppreference.com/w/cpp/language/copy_constructor

But there's no harm in explicitly deleting it, and it helps document the intent.

What's the motivation behind this?

The committee probably observed that in most cases where people wanted a move constructor, they needed either no copy constructor or a more complicated one (to handle duplicating a resource.) So creating a default copy constructor (a convenience in the common case of a simple structure) wasn't a win.

Perhaps if you were designing C++ from scratch without legacy and had a more Python-like mindset, you'd require an explicit "= default" rather than complicated rules about whether a default is provided for you or not.

The implicitly-declared copy constructor is deleted if there is a user-defined move constructor. See: https://en.cppreference.com/w/cpp/language/copy_constructor#...

So if you define a move constructor and still want your class to have a copy constructor, you'll have to explicitly define the copy constructor, too.

Luckily, in modern C++, defining the copy constructor in this case (where you just want to make the default available) is as simple as:

    A::A(const A &other) = default;

That makes sense, and hence the "rule of 5" (if you don't want it implicitly deleted). Thanks.

(author here), it is true, try it. Sample code is on https://github.com/ahupowerdns/hello-cpp/blob/master/move.cc

Thanks, and sorry for not just trying it. I'm on mobile and our workstations here aren't cpp11 capable.