zaxis 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)
>
> I know myFoldl implements foldl using foldr. However i really donot know
> how it can do it ?
>
> Please shed a light one me, thanks!
>
Hi,
Nice example! Well this is indeed an abstract piece of code. But basically
foldl f z xs starts with z and keeps applying (`f`x) to it
so for example foldl f z [1,2,3] = ((`f`3).(`f`2).(`f`1)) z
Because functions are first-class in haskell, we can also perform a foldl
where instead of calculating the intermediate values we calculate the
total function, i.e. ((`f`3).(`f`2).(`f`1)) and apply it to z.
When the list is empty z goes to z, so the start function must be id.
So we can write
(`f`3).(`f`2).(`f`1) = foldr (\x g -> g . (`f`x)) id xs
This is almost in your form.
Hope this helps,
Gerben
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