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:[EMAIL PROTECTED]
Sent: 18 November 2004 16:32
To: [EMAIL PROTECTED]
Cc: [EMAIL PROTECTED]
Subject: Re: [Hugs-users] Avoiding use of the stack
On 2004 November 18 Thursday 06:15, [EMAIL PROTECTED] 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.
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