CODEWITHSUNDEEP
listhaskell
Alan Kantserov
Alan Kantserov

Reputation: 37

lastList function in a list without calling a Prelude in Haskell

I have this function lastList that is suppose to call the last element of a list. I need to do manually, without predefined "last" in Prelude. I have this code so far

lastList :: List a -> a
lastList list = case list of
    Empty -> error "List is empty"
    Cons _ xs -> lastList xs

It gives me an output "List is empty" with the list 

test3 :: List Double
test3 = Cons 1.3 (Cons 5.2 (Cons 3.0 (Cons 8.7 (Cons 9.1 Empty))))

What is the problem? Thank you

Upvotes: 1

Views: 111

Answers (1)

willeM_ Van Onsem
willeM_ Van Onsem

Reputation: 477437

The problem: you only define Empty as basecase, which will error. So every non-infinite list, will eventually result in the Empty basecase, and error. Infinite lists will loop forever, but that is something we can not circumvent since infinite lists simply have no last element.

Since the List is a linked list with a head (which is an element) and a tail (which is the rest of the list) as Cons head tail and a Empty, a list with one element is thus Cons x Empty. So that means in case we obtain such a pattern, we should return x.

Furthermore in case the tail is not Empty, we should perform recursion on the tail, until we obtain a list with only one element left.

In case the list is Empty, the list is empty, and thus we should error on this input.

lastList :: List a -> a
lastList Empty = error "List is empty"
lastList (Cons x Empty) = x
lastList (Cons _ xs) = lastList xs

Upvotes: 6

Related Questions