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

Xingzhi Pan wrote:

More over, does "foldr step f id xs z" equal to "foldr (step f) id xs z"??

No, it does not. The former passes three-argument function 'step' to foldr and the later passes two-argument function which is the result of the partial application (step f). -- View this message in context: http://old.nabble.com/foldl-in-terms-of-foldr-tp27322307p27325448.html Sent from the Haskell - Haskell-Cafe mailing list archive at Nabble.com.