How to enforce copy elision, why it won't work with deleted copy constructor?

问题

I have an uncopiable class. Copying this would be problematic. I want to guarantee that it won't be ever copied, so I made its copy constructor deleted:

class A {
  public:
    A();
    A(const A&) = delete;
};

A fun() {
  return A();
};

int main() {
  A a = fun();
};

Unfortunately, g++ won't compile this on the reason:

t.cc: In function ‘A fun()’:
t.cc:8:12: error: use of deleted function ‘A::A(const A&)’
   return A();
            ^
t.cc:4:5: note: declared here
     A(const A&) = delete;
     ^
t.cc: In function ‘int main()’:
t.cc:12:13: error: use of deleted function ‘A::A(const A&)’
   A a = fun();
             ^
t.cc:4:5: note: declared here
     A(const A&) = delete;
     ^

But this is a very clear situation where copy elision should be used, so the copy constructor shouldn't be ever called. Why is it so?

回答1:

Until C++17 copy elision is an optimization the compiler is not required to do, so classes must be copyable since the compiler might want to copy (even if it actually does not). In C++17 copy elision will be guaranteed in many cases and then classes won't need copy ctors.

See also:

http://en.cppreference.com/w/cpp/language/copy_elision

http://www.open-std.org/jtc1/sc22/wg21/docs/papers/2015/p0135r0.html

https://herbsutter.com/2016/06/30/trip-report-summer-iso-c-standards-meeting-oulu/ (the bit about "Guaranteed copy elision")

You could perhaps use the old trick of declaring the copy constructor in your class but not actually implement it? That should please the compiler as long as it does not actually invoke the copy ctor. I didn't test that, but I believe it should work for your case until C++17 arrives.

回答2:

You can't force copy elision (yet) (see other answers).

However, you can provide a default move constructor for your class, this will move (and thus, not copy) the return value if RVO/NRVO is not possible. To do this you should add = default for your move constructors:

class A {
  public:
    A() = default;
    A(const A&) = delete;
    A(A&&) = default;
    A& operator=(A&&) = default;
};

Example

回答3:

Return value optimization (RVO and NRVO) does not mean the requirement that the types involved by copyable or movable is dropped. This requirement applies, whether you get RVO or not.

The most likely reason this is so is that copy elision is not (currently) enforced. It is an optimization that may take place, and it would not make sense for code to compile or not based on whether that optimization is applied in a particular implementation.

In C++17, RVO wil be enforced in some circumstances, and the requirements of copyability and movability will be dropped.

来源：https://stackoverflow.com/questions/49893051/c-object-array-initialization-without-copy-constructor