Catch empty list exeption in Haskell (head)-CodePudding

I'm writing a function that gets the index of the first even number from a list. The list I get may or may not contain even numbers, and I'd like to return -1 if there are no even numbers in the list. The list can be infinite.

I wrote this

posicPrimerPar'' :: [Int] -> Int
posicPrimerPar'' a = fromJust (elemIndex (head (filter (even) a)) a)

I could do something like:

posicPrimerPar' :: [Int] -> Int
posicPrimerPar' a = case length evens of
  0 -> -1;
  n -> fromJust elemIndex (head evens) a
  where evens = filter (even) a

But as you can see, this is not the most efficient way of doing it. A list [1..100000] contains a lot of even numbers, and I just need the first one. I need Haskell's laziness, so I need to ask for the head right there, but head throws an empty list exception when the list is empty (i.e. there are no even numbers in the list). I cannot find the Haskell equivalent of Python's try: ... except: .... All I could find regarding exceptions were IO related. What I need is except Prelude.head = -1 or something like that.

CodePudding user response：

Haskell is lazy, so evens will not be fully evaluated. The problematic part is the length evens which is not necessary. You can check with null :: Foldable f => f a -> Bool, or with pattern matching. For example:

import Data.List(findIndex)

posicPrimerPar' :: [Int] -> Maybe Int
posicPrimerPar' [] = Nothing
posicPrimerPar' xs = findIndex even xs

for findIndex :: (a -> Bool) -> [a] -> Maybe Int, you however do not need to take into account the empty list, since it already considers this.

or we can return -1 in case there is no such item:

import Data.List(findIndex)
import Data.Maybe(fromMaybe)

posicPrimerPar' :: [Int] -> Int
posicPrimerPar' = fromMaybe (-1) . findIndex even