How will the NRV optimization effect r-values?

[James Smith]

NRV changes a function that processes and returns an object into a
function that takes a reference to that object. When using r-values
this optimization seems to be unnecessary. I assume any code that
requires the optimization would now use r-values instead.


James Smith
(e-mail address removed)

 

[Alf P. Steinbach]

* James Smith:

NRV changes a function that processes and returns an object into a
function that takes a reference to that object. When using r-values
this optimization seems to be unnecessary. I assume any code that
requires the optimization would now use r-values instead.

Probably you're talking about rvalue references.

No, what goes on at the machine code level doesn't change.

It's how it's expressed at the C++ level, and what guarantees you have, that
changes.


Cheers & hth.,

- Alf

 

[Noah Roberts]

* James Smith:

Probably you're talking about rvalue references.

No, what goes on at the machine code level doesn't change.

It's how it's expressed at the C++ level, and what guarantees you have, that
changes.

The VS2010RC docs/websites claim that if NRV is not possible it will use
the move constructor if it is available. I don't know if this is a
requirement of C++0x or not.