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 -- View this message in context: http://www.nabble.com/Would-you-mind-explain-such-a-code---tp25377949p253782... Sent from the Haskell - Haskell-Cafe mailing list archive at Nabble.com.