[Haskell-cafe] foldl in terms of foldr

[Xingzhi Pan]

On Wed, Jan 27, 2010 at 12:05 AM, Eduard Sergeev
<Eduard.Sergeev at gmail.com> wrote:
>
>
> 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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-- 
Pan, Xingzhi
http://www.panxingzhi.net