On Dec 18, 2:20 pm, "Alex Blekhman" <tkfx.REM...@yahoo.com> wrote:
"Abhishek Padmanabh" wrote:
So, instead of making redundant copy of a string, you
just
bind a reference to return value of `foo' and use it as
long
as you need it.
I think what you are pointing towards is optimization
rather than
convenience. And if it is optimization that you are
hinting towards
then is that not what RVO/NRVO would take care of in this
case?
You can see it both ways. Consider following code:
struct X
{
X(int) { ... }
...
};
const X& result = X(1) + X(2);
Is it optimization or convenience? I could write it:
Optimization, of course.
X result = X(1) + X(2);
However, redundant copy would be created.
Yes, and that redundant copy is what can be optimized away by return
value optimization. You could have written either but there is a
difference in the way you could use the returned value of the right
hand operation. For first one, you cannot modify the result later on
although it does live but for the second, from performance point of
view it is same (with rvo), but you have an object that you can modify
- work with. It is not const. However if you did:
const X result = X(1) + X(2);
it would now be same. I do not see the convenience element there
though. My point is - temporary being able to be bound to a reference
to const does not root out of convenience or optimization (may be
optimization before compilers became capable of RVO - I won't know
much about history of compilers, the evolution).