CODEWITHSUNDEEP
haskell
dagda1
dagda1

Reputation: 28760

foldl expression not compiling due to use of literal

I have the following use of foldl which is erroring:

elementAt :: [a] -> Int -> a
elementAt [] x = error "No empty lists for element-at"
elementAt xs x = foldl(\acc (a, b) -> if(b == x) then a else acc) 0 $ zip xs [0..]

When I try and compile I get this error:

exercises.hs:8:67: error: * No instance for (Num a) arising from the literal 0' Possible fix: add (Num a) to the context of the type signature for: elementAt :: [a] -> Int -> a * In the second argument offoldl', namely `0' In the expression: foldl (\ acc (a, b) -> if (b == x) then a else acc) 0 In the expression: foldl (\ acc (a, b) -> if (b == x) then a else acc) 0 $ zip xs [0 .. ] Failed, modules loaded: none.

Upvotes: 0

Views: 77

Answers (1)

sepp2k
sepp2k

Reputation: 370082

When you don't find the element, the result of your function is 0. That only makes sense if you're working with a list of numbers. If you pass in a list of strings and then return either a string from the list or the number 0, that'd be a clear type error.

So your function only works with lists of numbers and your type signature must reflect that by adding a Num a constraint.

However a better solution would be to not use 0 as the default value and use a Maybe instead. That way you don't have to restrict yourself to lists of numbers.

Upvotes: 5

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