Hi Scott, Thanks for your insight on this one. My test routines now work, and I now have a better understanding of this aspect of Haskell. Best regards, Eoin -----Original Message----- From: Scott Turner [mailto:p.turner@computer.org] Sent: 18 November 2004 16:32 To: hugs-users@haskell.org Cc: eoin.mcdonnell@lineone.net Subject: Re: [Hugs-users] Avoiding use of the stack On 2004 November 18 Thursday 06:15, eoin.mcdonnell@lineone.net wrote:

how to write recursive functions so that they don't eat up stack space, and instead behave more like while loops (I think you had to convert them to tail-recursive forms?).

testR2 :: Integer -> Integer testR2 n = testR2' 0 n testR2' :: Integer -> Integer -> Integer testR2' a 0 = a testR2' a n = testR2' (a+n) (n-1)

You're on the right track. This is properly tail recursive. The remaining problem is that the argument (a+n) does not get evaluated during the processing of testR2'. So each successive call to testR2' is passed a larger expression such as testR2' (((((0+10)+9)+8)+7)+6) (6-1) Pattern matching causes the second argument to be evaluated. To ensure that the first argument is processed, you can use the $! operator. testR2' a n = (testR2' $! (a+n)) (n-1) which evaluates (a+n) before passing it to testR2'. Where you see stack overflow, it's actually a stack of + opeations rather than a stack of testR2' calls. __________________________________________________________________ INTRODUCTORY OFFER! Tiscali Business Broadband for £15.99! http://www.tiscali-business.co.uk/broadband/?code=ZZ-MS-12KC