Question

I’m trying to delete all instances of an item in a list using haskell. I get an error that I don’t really understand. Can anyone help me out and let me know if I’m doing the correct thing?

deleteAllInstances :: (a, [l]) =>  a -> [l] -> [l]
deleteAllInstances (a, []) = []
deleteAllInstances (i, (x:xs))
    | i == x = tail
    | otherwise = x ++ tail
    where tail = deleteAllInstances i xs

Answer 1

First, the type signature is malformed.

deleteAllInstances :: (a, [l]) =>  a -> [l] -> [l]

A type signature has the form

name :: (Constraints) => type

where Constraints involve type classes, like (Ord a, Show a). In this case, the function uses (==), so there must be a constraint of the form Eq a.

Then the function definition doesn’t match the type part, you defined it to take a pair as argument, while the type signature says otherwise (your definition is uncurried, the type is curried).

deleteAllInstances (a, []) = []
deleteAllInstances (i, (x:xs))
    | i == x = tail
    | otherwise = x ++ tail
    where tail = deleteAllInstances i xs

then you use (++) to glue an element to the front of a list, but (++) concatenates two lists, you need (:) here.

The simplest way to define the function would be to use filter

deleteAllInstances :: Eq a => a -> [a] -> [a]
deleteAllInstances a xs = filter (/= a) xs

but if you want to do the explicit recursion yourself,

deleteAllInstances :: Eq a => a -> [a] -> [a]
deleteAllInstances a (x:xs)
    | a == x    = rest
    | otherwise = x : rest
      where
        rest = deleteAllInstances a xs
deleteAllInstances _ _ = []