[Haskell-cafe] scanr producer?

[Louis Wasserman]

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
wasserman.louis at gmail.com
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