Do algebraic datatypes in Haskell equal discriminated unions in F#?

Question

I am learning Haskell and would like to know whether the constructs known in Haskell as algebraic datatypes are the same that discriminated unions in F# or there are some subtle differences between them.

I would also appreciate much a good comparison between F# (my first functional language) and other functional languages, especially as regards similar concepts but with substantial but important differences.

In response to your second question, this answer lists some F# distinctives. — Daniel

newacct newacct · Accepted Answer · 2011-11-16T21:34:57

(I come from OCaml, but I looked over the relevant F# stuff and it seems the same. Correct me if I'm wrong.) They are the same, just different terminology for the same thing, but there are a few syntactical differences. For example, to define a constructor with multiple data elements, in OCaml and F# you write the type as if they were stuffed in a tuple:

Haskell:

data Whatever = Foo TypeA TypeB

OCaml / F#:

type whatever = Foo of typeA * typeB

Similarly, to pattern match on it, you similarly act like a single argument that is a tuple with all the data members stuffed inside:

Haskell:

case x of Foo a b -> ...

OCaml / F#:

match x with Foo (a, b) -> ...

Edit: apparently the following does not apply in F#

~~Also, in Haskell the constructor automatically becomes a function that you can use by itself like any other value:~~

zipWith Foo xs ys

OCaml/F# don't do this. You could manually define your own functions for each constructor.

Do algebraic datatypes in Haskell equal discriminated unions in F#?

2 Answers