Why does copy elision make an exception for formal parameters?

Question

Here is a complete program:

#include 
using std::cout;
using std::endl;
using std::move;

int count {0};  // global for monitoring


class Triple         


        

Answer 1

First of all, understand that RVO and NRVO are opportunities offered by the writers of the standard to writers of compilers. A given compiler is free to ignore the possibility of RVO or NRVO if it can't make it work, if it doesn't know if it can make it work, if the moon is full, etc.

In this case, though, it's easy. The way (N)RVO is fundamentally implemented, is by constructing the return value directly into the memory occupied by the return value, or even the memory occupied by the variable which will be set to that return value.

That is, the potential savings from (N)RVO come not just from the ability to elide copy construction, but the ability to reduce copying in general.

But when the source of the return value is a function parameter, it's too late for that. left is already somewhere in memory, and the return value has to go somewhere else. Short of some brute-force as-if-ruling from inlining, copying is already a given, because a second object has already been constructed.