[Haskell-cafe] foldl in terms of foldr

[Eduard Sergeev]

Neil Brown-7 wrote:
> 
> step is of type b -> (a -> a) -> (a -> a), which does agree with (a -> b 
> -> b)

Not quite right..
Let's rewite the function:

myFoldl f z xs = foldr (step f) id xs z
step f x g = \a -> g (f a x) 

now (from ghci):
step (+) :: (Num t1) => t1 -> (t1 -> t3) -> t1 -> t3

or even:
step (flip (:)) :: t -> ([t] -> t3) -> [t] -> t3

But yes, the type from my first post was wrong

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