Yes. f somedata1 somedata2 = aa where aa = AA somedata1 bb bb = BB somedata2 aa
Spasibo, Yevgeny! Originally I was thinking theoretically about a single plain lambda-expression, like (\ somedata1 somedata2 -> (\ aa bb -> aa (bb aa)) (\ b -> AA somedata1 b) (\ a -> BB somedata2 a) ) But in the code "aa (bb aa)" last aa stays lacking an argument, of course, if we don't consider 1st application "aa (" as having a side effect on aa. And that's where "separate and rule" shows up it's power (speaking about "where" and namespacing in general). =) Belka -- View this message in context: http://www.nabble.com/Recursive-referencing-tp21722002p21722503.html Sent from the Haskell - Haskell-Cafe mailing list archive at Nabble.com.