[PATCH] revset: fix infinite recursion if called with __len__ first

[Pierre-Yves David]

On 03/31/2016 07:44 AM, Maciej Fijalkowski wrote:
> # HG changeset patch
> # User Maciej Fijalkowski <fijall at gmail.com>
> # Date 1459435465 -7200
> #      Thu Mar 31 16:44:25 2016 +0200
> # Node ID eaaeff1b98571b90ec71614c75126e64e98c004b
> # Parent  ff0d3b6b287f89594bd8d0308fe2810d2a18ea01
> revset: fix infinite recursion if called with __len__ first
>
> The problem here is simple - set(x) can (but doesn't have to) call
> __len__ first. If we do that, __len__ calls set which calls __len__
> again and we end up with inifinite recursion that gets swallowed and
> gets interesting results. Call it properly on the subset instead,
> fixes test-bisect2.t on PyPy
>
> diff -r ff0d3b6b287f -r eaaeff1b9857 mercurial/revset.py
> --- a/mercurial/revset.py	Tue Mar 29 12:29:00 2016 -0500
> +++ b/mercurial/revset.py	Thu Mar 31 16:44:25 2016 +0200
> @@ -3020,7 +3020,7 @@
>
>       def __len__(self):
>           # Basic implementation to be changed in future patches.
> -        l = baseset([r for r in self])
> +        l = baseset([r for r in self._subset if self._condition(r)])

yikes, duplicating the logic for an obscure reason seems suboptimal. 
Moreover, without a comment, the next guy wil factorise this :-/


What about

   baseset([r for r in iter(self)])

or maybe even "better".

   l = 0
   for r in self:
      l += 1
   return l

-- 
Pierre-Yves David