[Haskell-beginners] Function result being inversed

[Xavier Shay]

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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