Return local value from function without triggering copy constructor

Question

I am attempting to delete the copy constructor using the c++ type system to prevent copying an object.

struct DeleteCopyConstructor {
    DeleteCopyConstructor() {};
    DeleteCopyConstructor(DeleteCopyConstructor& op2) = delete;
    DeleteCopyConstructor(const DeleteCopyConstructor& op2) = delete;
};

DeleteCopyConstructor f() {
    DeleteCopyConstructor d;
    // initialize d...
    return d;
}

The error is:

error: use of deleted function ‘DeleteCopyConstructor::DeleteCopyConstructor(const DeleteCopyConstructor&)’

I've read about copy elision, but it appears to be a compiler optimization, so I don't think it applies. How can I return d without triggering copy construction?

Answer 1

In C++17 the compiler is guaranteed to elide the copy. However in all cases you still need to have a valid copy constructor.

So if you're just concerned about performance then you don't have to do anything. If you want to delete the copy constructor because the value logically shouldn't be copyable then there is no way to do it as far as I know. You'd have to return a std::unique_ptr<T> or take the value by reference and move into it that way. Edit: Or define a move constructor.