[Haskell-cafe] Would you mind explain such a code ?

[staafmeister]

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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