Why are Haskell algebraic data types “closed”?

Question

Correct me if I\'m wrong, but it seems like algebraic data types in Haskell are useful in many of the cases where you would use classes and inheritance in OO languages. But

Answer 1

The fact that ADT are closed makes it a lot easier to write total functions. That are functions that always produce a result, for all possible values of its type, eg.

maybeToList :: Maybe a -> [a]
maybeToList Nothing  = []
maybeToList (Just x) = [x]

If Maybe were open, someone could add a extra constructor and the maybeToList function would suddenly break.

In OO this isn't an issue, when you're using inheritance to extend a type, because when you call a function for which there is no specific overload, it can just use the implementation for a superclass. I.e., you can call printPerson(Person p) just fine with a Student object if Student is a subclass of Person.

In Haskell, you would usually use encapsulation and type classes when you need to extent your types. For example:

class Eq a where
   (==) :: a -> a -> Bool

instance Eq Bool where
  False == False = True
  False == True  = False
  True  == False = False
  True  == True  = True

instance Eq a => Eq [a] where
  []     == []     = True
  (x:xs) == (y:ys) = x == y && xs == ys
  _      == _      = False

Now, the == function is completely open, you can add your own types by making it an instance of the Eq class.

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Note that there has been work on the idea of extensible datatypes, but that is definitely not part of Haskell yet.