Eq => function in Haskell

Question

I am trying to figure out how to write this function with the Eq function in Haskell.

An easy function that I am trying to implement is:

f :: Eq a =>          


        

Answer 1

You can also use span and a recursive call to make it work:

f :: Eq a => [a] -> [[a]]
f [] = []
f l@(x:xs) = grouped : f remainder
    where
        (grouped, remainder) = span (== x) l

Here you have the live example

Answer 2

The presence of the Eq typeclass in your type is a red herring: it is not related to the error you report. The error you report happens because you have defined how the function should behave when the list is empty (f [] =) and when the list has at least two elements (f (x:x':xs) =), but not when the list has exactly one element. The solution will be to add a case that begins*

f [x] = ????

and decide how to deal with one-element lists. Or to find some other way to write the function that deals with all of its cases.

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* Note that f [x] is the same thing as f (x:[]).