Re: [Haskell-cafe] foldl in terms of foldr

Am Dienstag 26 Januar 2010 16:44:11 schrieb Xingzhi Pan:

On Tue, Jan 26, 2010 at 11:24 PM, Eduard Sergeev

wrote:

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Xingzhi Pan wrote:

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The first argument to foldr is of type (a -> b -> a), which takes 2 arguments.  But 'step' here is defined as a function taking 3 arguments.  What am I missing here?

You can think of step as a function of two arguments which returns a function with one argument (although in reality, as any curried function, 'step' is _one_ argument function anyway): step :: b -> (a -> c) -> (b -> c)

e.g. 'step' could have been defined as such: step x g = \a -> g (f a x)

to save on lambda 'a' was moved to argument list.

Right.  But then step is of the type "b -> (a -> c) -> (b -> c)".  But as the first argument to foldr, does it agree with (a -> b -> a), which was what I saw when I type ":t foldr" in ghci?

No, typo by Eduard, step :: b -> (a -> c) -> (a -> c). foldr :: (t -> u -> u) -> u -> [t] -> u , so t === b, u === a -> c Now, in "foldr step id xs", id has type u === a -> c, hence c === a.