RefRange

[Jonathan M Davis]

On Sunday, August 26, 2012 10:17:13 Jonathan M Davis wrote:
> On Sunday, August 26, 2012 17:41:45 David wrote:
> > It's a RefRange, but not completly ... Can somebody explain me that
> > behaviour?
> > 
> > http://dpaste.dzfl.pl/643de2a3
> 
> refRange simply returns the original range if it's an input range rather
> than a forward range, since normally, when you have an input range, it
> isn't a value type, and there's no way to copy it, so operating on one
> reference of it is already the same as operating on all of them, making
> RefRange pointless.
> 
> However, you've done the odd thing of declaring a value type input range. I
> don't know why that would ever be done except through ignorance of how
> ranges work. So, refRange is actually returning a copy in your case, which
> is why you're having problems.

Though the fact that you ran into this issue may indicate that having refRange 
return the original if it isn't a forward range was a bad decision. I don't 
know. In the normal case, it's definitely better, because it avoids an 
unnecessary wrapper, but obviously, people can make mistakes. You should still 
be able use RefRange with an input range though, as long as you use it 
directly.

auto wrapped = RefRange!TestRange(&orig);

But it would be better IMHO to just fix it so that your range is a forward 
range, since there's no reason for it not to be.

- Jonathan M Davis