Re: [Haskell-cafe] foldl in terms of foldr

Xingzhi Pan wrote:

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?

You can think of step as a function of two arguments which returns a function with one argument (although in reality, as any curried function, 'step' is _one_ argument function anyway): step :: b -> (a -> c) -> (b -> c) e.g. 'step' could have been defined as such: step x g = \a -> g (f a x) to save on lambda 'a' was moved to argument list. -- View this message in context: http://old.nabble.com/foldl-in-terms-of-foldr-tp27322307p27324376.html Sent from the Haskell - Haskell-Cafe mailing list archive at Nabble.com.