or takes two values arguments haskell

Question

I am a beginner in haskell and I tried to define a simple function for Sieve of Eratosthenes but it says error:

    • Couldn't match expected type ‘Bool -> Bool’
                  with actual type ‘Bool’
    • The function ‘or’ is applied to two value arguments,
        but its type ‘t Bool -> Bool’ has only one
      In the expression: or (mod n x) (divl l x)
      In an equation for ‘divl’: divl (n : l) x = or (mod n x) (divl l x)
   |
13 | divl (n:l) x = or (mod n x) (divl l x)

erat l [] = l
erat l (x:t) = if divl l x then erat l t else erat (x:t) l

divl (n:l) x = or (mod n x) (divl l x)
divl [] x = True

I tried to write that as an operator with "`" but nothing worked

Answer 1

or is not the boolean OR operator; that's (||). or takes a list (well, Foldable) of Boolean values and returns True if at least one value is True.

> or []
False
> or [True, False]
True

(Note that or [] is defined to be False to preserve the identity or xs || or ys == or (xs ++ ys). As a concrete example, or [] || or [False] == or [False].)

Answer 2

Try to use (||) :: Bool -> Bool -> Bool instead of or :: Foldable t => t Bool -> Bool

https://hackage.haskell.org/package/base-4.17.0.0/docs/Prelude.html#v:-124--124-

https://hackage.haskell.org/package/base-4.17.0.0/docs/Prelude.html#v:or

i.e. mod n x || divl l x or or [mod n x, divl l x].

Answer 3

You need to check if the modulo is zero, or one of the remaining, you use (||) :: Bool -> Bool -> Bool:

divl (n:l) x = mod n x == 0 || divl l x
divl [] x = False

The basecase should be False, otherwise it will always return True. It is probably better to work with any however:

divl ls x = any ((0 ==) . (`mod` x)) ls

Answer 4

at the end summarizing the answers i resolved

erat :: [Int] -> [Int] -> [Int]
erat l [] = l
erat l (x:t) = if divl l x then erat l t else erat (x:l) t

divl (n:l) x = ((mod x n) == 0) || (divl l x)
divl [] x = False