Hmm. Your example relies on inlining 'arr' at its call site.
My guess is that you aren't using -O. In that case, there's no
cross-module inlining, so 'arr' doesn't get inlined.
Is that it?
Simon
> -----Original Message-----
> From: Kevin Atkinson
> Sent: Tuesday, June 29, 1999 4:44 AM
> To: [EMAIL PROTECTED]
> Subject: Rule question.
>
>
> I was playing around with rules and I got some unexpected results. By
> just moving some stuff into a separate module I got different results.
>
> Could some one explain to me while this program:
>
> module Main where
>
> data Arr ix el = Arr Int [(ix,el)] deriving Show
>
> replaceMany :: [(ix,el)] -> Arr ix el -> Arr ix el
> replaceMany = error "In Replace Many"
>
> {-# RULES
> "rule1" forall f,l,a. replaceMany (map f l) a =
> replaceManyMap f l
> a
> #-}
>
> replaceManyMap :: (v -> (ix,el)) -> [(ix,el)] -> Arr ix el -> Arr ix
> el
> replaceManyMap = error "In Replace Many Map"
>
> arr s l =
> let a = Arr s [] in
> replaceMany l a
>
> arr2 s l =
> arr s (map (\(i,e)->(i+2,e)) l)
>
> main = print$ arr2 10 [(1::Int,10::Int),(2,20)]
>
> produces "Fail: In Replace Many Map" as expected but:
>
> module Main where
>
> import T2
>
> arr2 s l =
> arr s (map (\(i,e)->(i+2,e)) l)
>
> main = print$ arr2 10 [(1::Int,10::Int),(2,20)]
>
> module T2 where
>
> data Arr ix el = Arr Int [(ix,el)] deriving Show
>
> replaceMany :: [(ix,el)] -> Arr ix el -> Arr ix el
> replaceMany = error "In Replace Many"
>
> {-# RULES
> "rule1" forall f,l,a. replaceMany (map f l) a =
> replaceManyMap f l
> a
> #-}
>
> replaceManyMap :: (v -> (ix,el)) -> [(ix,el)] -> Arr ix el -> Arr ix
> el
> replaceManyMap = error "In Replace Many Map"
>
> arr s l =
> let a = Arr s [] in
> replaceMany l a
>
> produces "Fail: In Replace Many"?
>
> And how to fix the problem.
>
> Thanks in advance.
>
> --
> Kevin Atkinson
> [EMAIL PROTECTED]
> http://metalab.unc.edu/kevina/
>
