GHC does a very simple form of CSE. If it sees let x = e in ....e.... it replaces the inner 'e' by x. But that's all at the moment. Simon | -----Original Message----- | From: glasgow-haskell-users-bounces@haskell.org [mailto:glasgow-haskell-users-bounces@haskell.org] | On Behalf Of Lennart Augustsson | Sent: 27 November 2006 13:40 | To: Christian Maeder | Cc: GHC Users Mailing List | Subject: Re: Common subexpression elemination (CSE) | | GHC doesn't normally do CSE. CSE can cause space leaks, so you can't | do it willy-nilly. | I'm sure there are some strict contexts where it could be done | safely, but I don't think ghc uses that information (yet). | | -- Lennart | | On Nov 27, 2006, at 08:34 , Christian Maeder wrote: | | > the following code does not run as fast as expected: | > | > modexp a e n = if e <= 0 then 1 else | > if even e | > then mod (modexp a (div e 2) n * modexp a (div e 2) n) n | > else mod (a * modexp a (e - 1) n) n | > | > it gets only fast if written as: | > | > modexp2 a e n = if e <= 0 then 1 else | > if even e | > then let d = modexp2 a (div e 2) n in mod (d * d) n | > else mod (a * modexp2 a (e - 1) n) n | > | > I wonder, why the common subexpression "modexp a (div e 2) n" is not | > eliminated in the first case. Does CSE work at all? | > | > For testing the exponent "e" should have at least 10 digits. | > | > Cheers Christian | > | > P.S. Other alternatives are slower, too | > | > modexp1 a e n = if e <= 0 then 1 else | > mod (a * modexp1 a (e - 1) n) n | > | > modexp3 a e n = mod (a ^ e) n | > _______________________________________________ | > Glasgow-haskell-users mailing list | > Glasgow-haskell-users@haskell.org | > http://www.haskell.org/mailman/listinfo/glasgow-haskell-users | | _______________________________________________ | Glasgow-haskell-users mailing list | Glasgow-haskell-users@haskell.org | http://www.haskell.org/mailman/listinfo/glasgow-haskell-users