CODEWITHSUNDEEP
haskell
wawaloo_17
wawaloo_17

Reputation: 157

Checking to see if an element exists in a list in Haskell

I'm trying to create a recursive function that checks to see whether or not "a" is in an array. I'm trying to take the head of the tail and compare it to "a". However, it doesn't return the correct boolean value every time because it only take the head of the tail once. how do i make this recursive?

isElement :: Eq a => a -> [a] -> Bool
isElement a [] = False
isElement a (x:xs)
  | a == x = True
  | otherwise = False
  where x = head(xs)

Upvotes: 2

Views: 7090

Answers (1)

Silvio Mayolo
Silvio Mayolo

Reputation: 70257

There are a couple of issues. First off, your where block is not accomplishing anything. x is already known to be the head of the list because it's part of the (x:xs) construct, which destructures the list directly into the variables. So that line can be entirely removed. Second, there's no actual recursion here. You need to call isElement in the false case to check the rest of the list, rather than simply returning False.

isElement :: Eq a => a -> [a] -> Bool
isElement a [] = False
isElement a (x:xs)
  | a == x = True
  | otherwise = isElement a xs

Note that compiling your program with -Wall will give you warnings about a lot of things, including unused variables. In particular, the fact that, in your example (after removing the where block), xs was an unused variable should be a red flag and, in this case, gives us some insight into the problem: you never used the rest of the list, so the rest of the list gets ignored.

Upvotes: 4

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