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.