Using std::move() when returning a value from a function to avoid to copy

Question

Consider a type T supporting the default move semantics. Also consider the function below:

T f() {
   T t;
   return t;
}

T o = f();

In th

Answer 1

Ok, I would like to drop a comment on this. This question (and the answer) made me believe that it is not necessary to specify std::move on the return statement. However I was just thought a different lesson while dealing with my code.

So, I have a function (it's actually a specialization) that takes a temporary and just returns it. (The general function template does other stuff, but the specialization does the identity operation).

template<>
struct CreateLeaf< A >
{
  typedef A Leaf_t;
  inline static
  Leaf_t make( A &&a) { 
    return a;
  }
};

Now, this version calls the copy constructor of A upon returning. If I change the return statement to

Leaf_t make( A &&a) { 
  return std::move(a);
}

Then the move constructor of A gets called and I can do some optimizations there.

It might not be 100% matching your question. But it is false to think that return std::move(..) is never necessary. I used to think so. Not any more ;-)