Is there a change in strictness with the following transformation? It seems
like scanr should be a good consumer and a good producer, intuitively.
scanrFB :: (a -> b -> b) -> b -> [a] -> (b -> c -> c) -> c -> c
scanrFB f x ls c n = snd (foldr (\ x (b, bs) -> let b' = f x b in (b', b'
`c` bs)) (x, n) ls)
{-# RULES
"scanr" [~1] forall f x l . scanr f x l = build (scanrFB f x l);
-# RULES}
--
Louis Wasserman
[EMAIL PROTECTED]
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