Question

This is a question about C++ specs on object destruction vs Return-Value-Optimization.

Can I expect RVO return the right value before std::unique_ptr<> cleanup?

Foo Bar()
{
    std::unique_ptr<Foo> ptr = new Foo();
    return *ptr;
}

Answer 1

It will return the right value with or without RVO (and there is no RVO in this case). The function returns a concrete Foo, so *ptr will be copied into the return value before destruction of the pointer.

That means,

Foo foo;
foo = Bar();

is similar to (unwrapping the unique_ptr to be more explicit)

  Foo foo;
  Foo* ptr = new Foo;
  foo = *ptr;
finally:
  delete ptr;