Is it possible to export constructors for pattern

问题:

A vanilla data type in Haskell has zero or more constructors, each of which plays two roles.

In expressions, it supports introduction, its a function from zero or more arguments to the data type.

In patterns, it supports elimination, its kinda like a function from the data type to Maybe (tuple of argument types).

Is it possible for a module signature to hide the former while exposing the latter?

The use case is this: I have a type, T, whose constructors types alone can sometimes be used to construct nonsense. I have construction functions which can be used to build instances of the type that are guaranteed not to be nonsense. It would make sense to hide the constructors in this case, but it would still be useful for callers to be able to pattern match over the guaranteed-non-nonsense that they build with the construction functions.

I suspect this is impossible, but in case anyone has a way to do it, I though I would ask.

Next best thing is to hide the constructors and create a bunch of functions from T -> Maybe (This, That), T -> Maybe (The, Other, Thing), etc.

回答1:

You can use a view type and view patterns to do what you want:

module ThingModule (Thing, ThingView(..), view) where

data Thing = Foo Thing | Bar Int

data ThingView = FooV Thing | BarV Int

view :: Thing -> ThingView
view (Foo x) = FooV x
view (Bar y) = BarV y

Note that ThingView is not a recursive data type: all the value constructors refer back to Thing. So now you can export the value constructors of ThingView and keep Thing abstract.

Use like this:

{-# LANGUAGE ViewPatterns #-}
module Main where

import ThingModule

doSomethingWithThing :: Thing -> Int
doSomethingWithThing(view -> FooV x) = doSomethingWithThing x
doSomethingWithThing(view -> BarV y) = y

The arrow notation stuff is GHC's View Patterns. Note that it requires a language pragma.

Of course you're not required to use view patterns, you can just do all the desugaring by hand:

doSomethingWithThing :: Thing -> Int
doSomethingWithThing = doIt . view
  where doIt (FooV x) = doSomethingWithThing x
        doIt (BarV y) = y

More

Actually we can do a little bit better: There is no reason to duplicate all the value constructors for both Thing and ThingView

module ThingModule (ThingView(..), Thing, view) where

   newtype Thing = T {view :: ThingView Thing}
   data ThingView a = Foo a | Bar Int

Continue useing it the same way as before, but now the pattern matches can use Foo and Bar.

{-# LANGUAGE ViewPatterns #-}
module Main where

import ThingModule

doSomethingWithThing :: Thing -> Int
doSomethingWithThing(view -> Foo x) = doSomethingWithThing x
doSomethingWithThing(view -> Bar y) = y

回答2:

From GHC 7.8 on you can use PatternSynonyms to export patterns independent from constructors. So an analogue to @Lambdageek’s answer would be

{-# LANGUAGE PatternSynonyms #-}

module ThingModule (Thing, pattern Foo, pattern Bar) where

pattern Foo a <- RealFoo a
pattern Bar a <- RealBar a

data Thing = RealFoo Thing | RealBar Int

and

{-# LANGUAGE PatternSynonyms #-}
module Main where

import ThingModule

doSomethingWithThing :: Thing -> Int
doSomethingWithThing (Foo x) = doSomethingWithThing x
doSomethingWithThing (Bar y) = y

So it looks like normal constructors.

If you try to use Bar to construct a value, you get

Main.hs:9:32:
    Bar used in an expression, but it's a non-bidirectional pattern synonym
    In the expression: Bar y

回答3:

You cannot. But if there are only reasonable number of constructors for your type T, you may want to hide the constructors and instead provide a function which does the pattern matching in the same spirit as maybe :: b -> (a -> b) -> Maybe a -> b.