[Haskell-cafe] foldl in terms of foldr
Ryan Ingram
ryani.spam at gmail.com
Tue Jan 26 15:56:49 EST 2010
On Tue, Jan 26, 2010 at 5:04 AM, Xingzhi Pan <vengeance.storm at gmail.com> wrote:
> myFoldl :: (a -> b -> a) -> a -> [b] -> a
> myFoldl f z xs = foldr step id xs z
> where step x g a = g (f a x)
>
> Btw, is there a way I can observe the type signature of 'step' in this code?
Here is how I do it:
Prelude> :t \[f] -> \x g a -> g (f a x)
\[f] -> \x g a -> g (f a x)
:: [t1 -> t -> t2] -> t -> (t2 -> t3) -> t1 -> t3
The [] are unnecessary but help me differentiate the "givens" from the
actual arguments to the function.
Here's a way that gives better type variables and properly constrains
the type of f:
Prelude> let test [f] x g a = g (f a x) where typeF = f `asTypeOf` const
Prelude> :t test
test :: [a -> b -> a] -> b -> (a -> t) -> a -> t
-- ryan
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