On Monday, 17 April 2017 at 18:07:36 UTC, Jonathan M Davis wrote:
In this particular case, it looks like the main problem is
RefRange's opAssign. For it to work, the type needs to be
copyable. It might be reasonable for RefRange to be enhanced so
that it doesn't compile in opAssign if the range isn't
copyable, but I'd have to study RefRange in depth to know what
the exact consequences of that would be, since it's been quite
a while since I did anything with it. My guess is that such a
change would be reasonable, but I don't know without studying
it.
- Jonathan M Davis
I took a look on RefRange and the reasoning is clearly explained
in the docs like this:
This does not assign the pointer of $(D rhs) to this $(D
RefRange).
Rather it assigns the range pointed to by $(D rhs) to the range
pointed
to by this $(D RefRange). This is because $(I any) operation on a
RefRange) is the same is if it occurred to the original range. The
exception is when a $(D RefRange) is assigned $(D null) either
or because $(D rhs) is $(D null). In that case, $(D RefRange)
longer refers to the original range but is $(D null).
But what I do not understand is why this is important.
