Пожалуйста :)
It's difficult to do this in lambda because it isn't expressible in
simply (or even polymorphically) typed lambda calculus, since it
requires the Y (fix) combinator, which introduces non-terminating
computations, whereas simply- and polymorphically-typed lambda
calcules are strongly normalizing (all computations terminate).
So, you can't write *simple and intuitive* typed code for this in lambda.
To express it, you either need untyped lambda or Haskell, where the
'fix' combinator is typed by kind of a hack :)
With the Y combinator, the code becomes as follows:
f = \somedata1 somedata2 -> fst $ fix (\(aa,bb) -> (AA somedata1 bb,
BB somedata2 aa))
I tried to prove to myself that the structure really turns out cyclic,
but games with reallyUnsafePtrEquality# didn't lead to success; that's
why it's "reallyUnsafe" :-/
2009/1/29 Belka
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.
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-- Евгений Кирпичев Разработчик Яндекс.Маркета