Haskell xor not work for mapping

user_4309230 0
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I have a problem of using xor function in Data.Bits module like a code below 

import Data.Bits

andFunc :: [Int] -> [Int] -> [Int]
andFunc xs ys = zipWith (\x y -> x .&. y) xs ys

xorFunc :: [Int] -> [Int] -> [Int]
xorFunc xs ys = zipWith (\x y -> x xor y) xs ys

When I try to apply andFunc with arguments of [1..10] and [2..11] (arguments are just arbitrary array)

it works. (Does not write here, but orFunc (.|.) also works)

but some reasons, xorFunc does not.... and says

<interactive>:74:1: error:
    ? Non type-variable argument
        in the constraint: Enum ((a -> a -> a) -> t -> c)
      (Use FlexibleContexts to permit this)
    ? When checking the inferred type
        it :: forall a t c.
              (Enum ((a -> a -> a) -> t -> c), Enum t,
               Num ((a -> a -> a) -> t -> c), Num t, Bits a) =>
              [c]

Do you know why?

Running Environment: GHC 8.2.1 with no flags Windows 10 64 bit

xor function haskell
3 Answers

If you want to use functions in infix notation you have to use backtick syntax.

xorFunc :: [Int] -> [Int] -> [Int]
xorFunc xs ys = zipWith (\x y -> x `xor` y) xs ys

but this can be solved a bit simpler by not writing this as a lambda expression

xorFunc :: [Int] -> [Int] -> [Int]
xorFunc xs ys = zipWith xor xs ys

and applying eta reduce (twice), i.e. omitting parameters that are occurring in the last position and can be fully derived by the type checker.

xorFunc :: [Int] -> [Int] -> [Int]
xorFunc = zipWith xor
user_6301600

Infix functions are spelled with punctuation and can be made prefix with parentheses; e.g. x + y can also be spelled (+) x y. Going the other direction, prefix functions are spelled with letters and can be made infix with backticks; e.g. zip xs ys can also be spelled xs `zip` ys.

Applying that to your case, this means you should write one of xor x y or x `xor` y instead of x xor y.

user_7916040

xor is a regular function name, not an operator. You need to enclose it in backquotes to use as an infix operator.

xorFunc xs ys = zipWith (\x y -> x `xor` y) xs ys

That said, your lambda expressions aren't necessary; just use xor as argument to zip:

xorFunc xs ys = zipWith xor xs ys

or simply

xorFunc = zipWith xor

(Likewise, andFunc = zipWith (.&.); enclose the operator in parentheses to use it as a function value.)

user_11268410
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