Use anonymous function as parameter

Question

So i currently have this function for generating all possible combinations

func :: [a] -> [b] -> [(a, b)]
func xs ys = concat(map (\x -> map (\y -> (x,y))ys) xs)

Now I want to extend it so that i can filter it using an anonymous function as parameter.

Example:

func [1,2,3] [4,5] (\ a b -> a+b > 6) -> [(2,5),(3,4),(3,5)]

I tried to change the Typesignature of func:

from: [a] -> [b] -> [(a, b)]

to: [a] -> [b] -> (a1 -> a1 -> Bool) -> t

But it did not work.

Thank you in advance!

Answer 1

OP has stated that they changed their code to

func :: [a] -> [b] -> ((a,b) -> Bool) -> [(a, b)]
func xs ys f = filter f (tup)
  where tup = concat(map (\x -> map (\y -> (x,y))ys) xs)

followed by what one assumes is

func [1,2,3] [4,5] (\a b -> a + b > 7)

which gives them the error

• Couldn't match expected type ‘Bool’ with actual type ‘(a, b) -> Bool’
• The lambda expression ‘\ a b -> a + b > 7’ has two value arguments, but its type ‘(a, b) -> Bool’ has only one

This is easy to fix; in fact the error message is relatively clear as GHC error messages go. (That's faint praise, admittedly.) Doing this instead should work, though I'm away from a Haskell compiler and can't run it:

func [1,2,3] [4,5] (\(a, b) -> a + b > 7)

Explanation: \a b -> a + b > 7 is an anonymous function taking two arguments, a and b, whereas \(a, b) -> a + b > 7 is an anonymous function taking a single argument which is a pair. The type of the last argument of func, ((a, b) -> Bool), is the type of the second anonymous function, not the first.

(This problem can also be fixed by making func's last argument be of type a -> b -> Bool, while passing the first anonymous function.)

I originally thought OP's tup was just zip, and recommended they use zip instead to improve legibility. OP correctly points out that their function has very little in common with zip except the type. There may still be a legibility win to be had, though; see this.