On Tuesday 25 January 2011 21:00:11, 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
It's because f (<) k = (f (<)) k = ((<) 4) k = (<) 4 k = 4 < k or, shorter, (<) 4 = (4 <)
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)
A right operator section, (<*> x) translates to flip (<*>) x in prefix notation.
Cheers, Xavier