Why is using the move constructor in a return statement legal?

Question

Consider the following:

#include 

#define trace(name) std::cout << #name << \" (\" << this << \"), i = \" << i         


        

Answer 1

C++11 12.8/32: When the criteria for elision of a copy operation are met or would be met save for the fact that the source object is a function parameter, and the object to be copied is designated by an lvalue, overload resolution to select the constructor for the copy is first performed as if the object were designated by an rvalue.

Returning a local automatic variable meets the criteria for elision; treating it as if it were an rvalue selects the move constructor.