It's not too difficult to do: think 'fold map' and put it in the form that foldr needs. cheers -----Original Message----- From: James Ealing [mailto:jamesealing2000@yahoo.co.uk] Sent: Friday, 02 January, 2004 1:54 PM To: haskell-cafe@haskell.org Subject: foldr Hi all, If I have a function: it f [a0, a1, a2, ...] = [a0, f a1, f (f a2), ...] Is there any way of expressing it f as an instance of foldr? Many thanks, Jim ________________________________________________________________________ Yahoo! Messenger - Communicate instantly..."Ping" your friends today! Download Messenger Now http://uk.messenger.yahoo.com/download/index.html _______________________________________________ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe