Question

If I want to call Bar() instead of Foo(), does Bar() return me a copy (additional overhead) of what Foo() returns, or it returns the same object which Foo() places on the temporary stack?

vector<int> Foo(){       vector<int> result;       result.push_back(1);       return result;   }   vector<int> Bar(){       return Foo();   } 

Answer 1

Both may happen. However, most compiler will not do copy as soon as you optimize.

Your code indicate there should be a copy. However, the compiler is allowed to remove any copy that do not change the semantic and the program.

Note: This is why you should NEVER have a copy constructor that does anything but copying correctly as you can never be sure if a copy will be actually done or not.