Thanks Neil. How do I add in another ~10 computations, or map a list of a
100 computations to the same input?
Isn't there a way to do this without one computation having to be aware of
the other?
This feels like a situation Parsec users would find themselves in all the
time. When you have a bunch of parsers in a 'choice', does the start of the
input stream linger until the last parser is executed?
Thanks,
Greg
On 3/27/06, Neil Mitchell
Hi,
Here would be a better example then.
f lst = show (sum (filter (> 1) lst), sum (filter (> 2) lst))
2 to p2 - to show how this can be done in the general case. With the specific information you know about >1 vs >2 you can do better, but
I suspected that you actually wanted to do something "cleverer" with the list, for the sake of argument, I'm going to change >1 to p1 and this gets across the general point:
f lst = show (sumPairs (>1) (>2) lst)
sumPairs :: (Int -> Bool) -> (Int -> Bool) -> [Int] -> (Int, Int) sumPairs p1 p2 [] = (0, 0) sumPairs p1 p2 (x:xs) = (add p1 a, add p2 b) where (a,b) = sumPairs xs add pred value = if pred x then x+value else value
[Untested, something like this should work]
You can actually arrive at this solution entirely be reasoning on the program, i.e. not coming up with a fresh definition.
The above code essentially follows your imperative pseudo code - I think its constant space, but I'm not too sure...
Thanks
Neil