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 answer has to do with in what ways the code is easy to extend, a tension between classes and algebraic data types that Phil Wadler dubbed "the expression problem":

• With algebraic data types,

  • It is very cheap to add a new operation on things: you just define a new function. All the old functions on those things continue to work unchanged.

  • It is very expensive to add a new kind of thing: you have to add a new constructor an existing data type, and you have to edit and recompile every function which uses that type.

• With classes,

  • It is very cheap to add a new kind of thing: just add a new subclass, and as needed you define specialized methods, in that class, for all the existing operations. The superclass and all the other subclasses continue to work unchanged.

  • It is very expensive to add a new operation on things: you have to add a new method declaration to the superclass and potentially add a method definition to every existing subclass. In practice, the burden varies depending on the method.

So, algebraic data types are closed because a closed type supports certain kinds of program evolution well. For example, if your data types define a language, it is easy to add new compiler passes without invalidating old ones or changing the data.

It is possible to have "open" data types, but except under carefully controlled circumstances, the type checking becomes difficult. Todd Millstein did some very beautiful work on a language design that supports open algebraic types and extensible functions, all with a modular type checker. I found his paper a great pleasure to read.