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

Eduard Sergeev wrote:

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).

Correction :) 4-arg and 3-arg respectively. -- View this message in context: http://old.nabble.com/foldl-in-terms-of-foldr-tp27322307p27325593.html Sent from the Haskell - Haskell-Cafe mailing list archive at Nabble.com.