[Haskell-cafe] foldl in terms of foldr

Xingzhi Pan vengeance.storm at gmail.com
Tue Jan 26 08:04:10 EST 2010


I am reading Real World Haskell and have some questions about the
piece of code implementing foldl in terms of foldr:

-- file: ch04/Fold.hs
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)

The first argument to foldr is of type (a -> b -> a), which takes 2
arguments.  But 'step' here is defined as a function taking 3
arguments.  What am I missing here?  (Partial application?  The order
of execution?)

Btw, is there a way I can observe the type signature of 'step' in this code?

Thanks in advance!

Pan, Xingzhi

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