On Wed, Jan 27, 2010 at 12:05 AM, 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)
I am not very sure about this. This rewriting was my first reaction against the original code but it failed compilation with GHC. More over, does "foldr step f id xs z" equal to "foldr (step f) id xs z"?? Thanks!
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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