Thanks to all responses. Makes sense now. For the record I have ended up with this: mySort [] = [] mySort (h:t) = f(<=) ++ [h] ++ f(>) where f x = mySort (filter (not . x h) t) I have a feeling I may be able to make the following work with some sort of type declaration but I haven't really learned much about them yet so I will revisit later. (At the moment it does not compile.) mySort [] = [] mySort (h:t) = f(<=) ++ [h] ++ f(>) where f x = mySort (filter (h x) t) Cheers, Xavier On 26/01/11 7:00 AM, Xavier Shay wrote:
Hello, I am confused by the following code. I would expect results of True, False.
$ ghci *Main> let f x = x 4 *Main> f(<) 3 False *Main> f(<) 5 True
This came about because I was trying to refactor a sort function I wrote:
mySort [] = [] mySort (h:t) = (f (<= h)) ++ [h] ++ (f (> h)) where f x = mySort (filter x t)
I came up with this, which appears to work, though the comparison operators are backwards.
mySort [] = [] mySort (h:t) = f(>) ++ [h] ++ f(<=) where f x = mySort (filter (x h) t)
Cheers, Xavier
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