Return a const reference or a copy in a getter function?

Question

What\'s better as default, to return a copy (1) or a reference (2) from a getter function?

class foo {
public:
    std::string str () { // (1)
              


        

Answer 1

The compiler will not be able to perform (N)RVO in this case. The (named) return value optimization is an optimization where the compiler creates the function auto variables in the place of the return value to avoid having to copy:

std::string f()
{
   std::string result;
   //...
   return result;
}

When the compiler sees the code above (and assuming that if any other return is present it will also return the result variable) it knows that the variable result has as only possible fate being copied over the returned temporary and then destroyed. The compiler can then remove the result variable altogether and use the return temporary as the only variable. I insist: the compiler does not remove the return temporary, it removes the local function variable. The return temporary is required to fulfill the compilers call convention.

When you are returning a member of your class, the member must exist, and the call convention requires the returned object to be in a particular location (stack address usually). The compiler cannot create the method attribute over the returned object location, nor can it elide making the copy.