Haskell: Non-exhaustive patterns in function (simple functions) [duplicate]

Answer 1

You need to understand the syntax better.

Basically, there are 2 sub-syntaxes:

1. syntax for types
2. syntax for expressions and patterns

In the type syntax, [a] means list of elements of type a

In the expression/pattern syntax, [a] means the singleton list, that contains the value a. This is equivalent to (a:[]) (a prepended to the empty list).

Hence your first function, for example, checks if it gets a singleton list. Then it takes the head of the tail of the tail of a singleton list, which will fail.

The message you're getting is because there are shapes of lists you didn't cover: namely, the empty list and lists with more than 1 element.

And, of course, you should be getting a warning for third2, is it only covers lists with 3 or more elements. I am sure you're overlooking something.

Answer 2

That is odd that the second one does not complain about non-exhaustive patterns, since third2 will not match lists of length zero, one, or two. The third and third3 functions complain because [a] is not a variable, it is a pattern. [a] desugars to (a:[]), so you could have written them as

third (a:[]) = head (tail (a:[]))

third3 (a:[]) = (a:[]) !! 2

Neither of which will work, as those are single element lists. I suspect what you want is

third a = head (tail a)

third3 a = a !! 2