Non-exhaustive patterns in function scalarNew

Question

What does "Non-exhaustive patterns in function scalarNew" mean in this context?

scalarNew :: [Integer]->[Integer]->Integer
scalarNew (x:xs) (y:ys) 
     | (length xs == length ys) = x * y + scalarNew xs ys
     | otherwise = error "error"

Try to write a version without recursion. Hint: use `zipWith` and `sum`. — Elmex80s, Jan 22 '19 at 13:57

lsmor · Accepted Answer · 2019-01-22T14:57:46.217

5

You have to add the base case when using recursion. In this situation, the base case is when both are empty lists.

scalarNew :: [Integer] -> [Integer] -> Integer
scalarNew []     []        = 0
scalarNew []     (y:ys)    = error "error"
scalarNew (x:xs)    []     = error "error"
scalarNew (x:xs) (y:ys) 
  | length xs == length ys = x * y + scalarNew xs ys
  | otherwise = error "error"

edit: handle only-one-empty-list case

edited Jan 22 '19 at 14:57

answered Jan 22 '19 at 13:27

lsmor

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ok, i understand , thank u – liu fish Jan 22 '19 at 13:31
don't forget to mark as solved ;) – lsmor Jan 22 '19 at 13:31
1

You've neglected two cases: when only one of the two lists are empty. This will continue to produce an error in some cases. – AJF Jan 22 '19 at 13:32
@AJFarmar yep! forgot it actually. Already edited, thanks ;) – lsmor Jan 22 '19 at 13:37

Non-exhaustive patterns in function scalarNew

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