Extending a datatype in Haskell

Question

Haskell newbie here.

I wrote an evaluator for a minimal assembly-like language.

Now, I want to extend that language to support some syntactic sugar which, I will then compile back to use only the primitive operators. The ideia is that I do not want to touch the evaluator module again.

In the OO way of doing things, I think, one could extend the original module so to support the syntactic sugar operators, providing here the translation rules.

Other than that, I can only think of rewriting the datatype constructors in both modules so that they would not name-collide, and proceed from there, as if they were complete different things, but that implies some redundancy, for I would have to repeat (just with other names) the operators in common. Again, I think the keyword here is extend.

Is there a functional way of accomplishing this?

Thanks for taking the time to read this question.

Answer 1

A (simpler) option is to add a type to your AST, to distinguish Core from Extended:

data Core = Core
data Extended = Extended

data Expr t 
  = Add (Expr t) (Expr t)
  | Mult (Expr t) (Expr t)
  | Const Int 
  | Sugar t (Expr t) (Expr t)

An expression is either Core or Extended: the compiler will ensure that it contains only sub-expressions of the same type.

The function signatures in your original module would need to use Expr Core (instead of just Expr)

A Desugar function would have the following type signature:

Desugar :: Expr Extended -> Expr Core

You may also be interested in the more sophisticated approach described in the paper 'Trees that grow'.

Answer 2

This problem was named "the expression problem" by Phil Wadler, in his words:

The goal is to define a data type by cases, where one can add new cases to the data type and new functions over the data type, without recompiling existing code, and while retaining static type safety.

One solution to have extensible data type is to use type classes.

As an example let's assume we have a simple language for arithmetics:

data Expr = Add Expr Expr | Mult Expr Expr | Const Int

run (Const x) = x
run (Add exp1 exp2)  = run exp1 + run exp2
run (Mult exp1 exp2) = run exp1 * run exp2

e.g.

ghci> run (Add (Mult (Const 1) (Const 3)) (Const 2))
5

If we wanted to implement it in an extensible way, we should switch to type classes:

class Expr a where
    run :: a -> Int


data Const = Const Int

instance Expr Const where
    run (Const x) = x


data Add a b = Add a b

instance (Expr a,Expr b) => Expr (Add a b) where
    run (Add expr1 expr2) = run expr1 + run expr2


data Mult a b = Mult a b

instance (Expr a, Expr b) => Expr (Mult a b) where
    run (Mult expr1 expr2) = run expr1 * run expr2

Now let's extend the language adding subtractions:

data Sub a b = Sub a b

instance (Expr a, Expr b) => Expr (Sub a b) where
    run (Sub expr1 expr2) = run expr1 - run expr2

e.g.

ghci> run (Add (Sub (Const 1) (Const 4)) (Const 2))
-1

For more info on this approach, and in general on the expression problem, check Ralf Laemmel's videos 1 and 2 on Channel 9.

However, as noticed in the comments, this solution changes the semantics. For example lists of expressions are no longer legal:

[Add (Const 1) (Const 5), Const 6] -- does not typecheck

A more general solution using coproducts of type signatures is presented in the functional pearl "Data types a la carte". See also Wadler's comment on the paper.

Answer 3

You could do something a bit more OOP-like using existential types:

-- We need to enable the ExistentialQuantification extension.
{-# LANGUAGE ExistentialQuantification #-}

-- I want to use const as a term in the language, so let's hide Prelude.const.
import Prelude hiding (const)

-- First we need a type class to represent an expression we can evaluate
class Eval a where
  eval :: a -> Int

-- Then we create an existential type that represents every member of Eval
data Exp = forall t. Eval t => Exp t

-- We want to be able to evaluate all expressions, so make Exp a member of Eval.
-- Since the Exp type is just a wrapper around "any value that can be evaluated,"
-- we simply unwrap that value and call eval on it.
instance Eval Exp where
  eval (Exp e) = eval e

-- Then we define our base language; constants, addition and multiplication.
data BaseExp = Const Int | Add Exp Exp | Mul Exp Exp

-- We make sure we can evaluate the language by making it a member of Eval.
instance Eval BaseExp where
  eval (Const n) = n
  eval (Add a b) = eval a + eval b
  eval (Mul a b) = eval a * eval b

-- In order to avoid having to clutter our expressions with Exp everywhere,
-- let's define a few smart constructors.
add x y = Exp $ Add x y
mul x y = Exp $ Mul x y
const   = Exp . Const

-- However, now we want subtraction too, so we create another type for those
-- expressions.
data SubExp = Sub Exp Exp

-- Then we make sure that we know how to evaluate subtraction.
instance Eval SubExp where
  eval (Sub a b) = eval a - eval b

-- Finally, create a smart constructor for sub too.
sub x y = Exp $ Sub x y

By doing this, we actually get a single extendable type so you could, for example, mix extended and base values in a list:

> map eval [sub (const 10) (const 3), add (const 1) (const 1)]
[7, 2]

However, since the only thing we now can know about Exp values is that they are somehow members of Eval, we can't pattern match or do anything else that isn't specified in the type class. In OOP terms, think of Exp an exp value as an object that implements the Eval interface. If you have an object of type ISomethingThatCanBeEvaluated, obviously you can't safely cast it into something more specific; the same applies to Exp.

Answer 4

Syntactic sugar is usually handled by a parser; you'd extend (not in the sense of OO inheritance) the parser to detect the new constructs and translate them to the kind of structures that your evaluator can handle.