Non exhaustive pattern in [String] ->String?

Question

so I\'m creating an function like this:

 unlinhas::[String]->String
 uninhas [x] = \"\"
 unlinhas (x:xs) = x ++ \"\\n\" ++(unlinhas( xs))

th

Answer 1

Your first pattern [x] is a list with one element. So the Haskell compiler is wondering what to do with an empty list.

Furthermore in your first line you write uninhas, instead of unlinhas, as a result the Haskell compiler thinks that you write two different functions.

Based on your specifications however, you want to process the empty list, so you can fix it with:

unlinhas:: [String] -> String
unlinhas [] = ""
unlinhas (x:xs) = x ++ "\n" ++(unlinhas( xs))

You can further cleanup the code, and write it as:

unlinhas:: [String] -> String
unlinhas [] = ""
unlinhas (x:xs) = x ++ '\n' : unlinhas xs