I have a function func in haskell of the form func :: (Foldable t, Eq b, Num b) => t b -> b, this is intended to take a function from a pre-existing main function (i.e. I can't change the input/output pattern), which is a helper of the form helper :: (Eq a,Num a) => a -> a -> a, as well as a list of values, and I want to fold the list with the operation given in the helper function.
The full code block is
helper :: (Eq a,Num a) => a -> a -> a
helper x y
| x == 0 && y == 0 = 0
| x == 0 && y == 1 = 1
| x == 1 && y == 0 = 1
| x == 1 && y == 1 = 0
func :: (Foldable t, Eq b, Num b) => t b -> b
func help lst = foldl (help) 0 lst
Which gives a compilation error:
File.hs:23:23: error:
• Couldn't match type ‘b’ with ‘a0 -> b0’
Expected: b0 -> a0 -> b0
Actual: t b
‘b’ is a rigid type variable bound by
the type signature for:
func :: forall (t :: * -> *) b.
(Foldable t, Eq b, Num b) =>
t b -> b
at File.hs:22:1-44
• In the first argument of ‘foldl’, namely ‘(help)’
In the expression: foldl (help) 0 lst
In an equation for ‘func’: func help lst = foldl (help) 0 lst
• Relevant bindings include
lst :: t0 a0 (bound at File.hs:23:10)
help :: t b (bound at File.hs:23:5)
func :: t b -> b (bound at File.hs:23:1)
I honestly have no idea how to fix this.
You use the helper function here, not a parameter, so you work with:
func :: (Foldable t, Eq b, Num b) => t b -> b
func lst = foldl helper 0 lastor shorter:
func :: (Foldable t, Eq b, Num b) => t b -> b
func = foldl helper 0You can also make use of xor :: Bits a => a -> a -> a:
import Data.Bits(Bits, xor)
func :: (Foldable t, Bits b, Num b) => t b -> b
func = foldl xor 0