[Haskell-cafe] exporting Constructors as pattern-only

[Jon Purdy]

I believe you can use unidirectional pattern synonyms, and only export the
patterns, not the constructors.

{-# LANGUAGE PatternSynonyms #-}
module Tee (T, t1, t2, pattern T1, pattern T2) where

data T = MkT1 Bool | MkT2 Int | …
pattern T1 a <- MkT1 a
pattern T2 a <- MkT2 a
…
t1 :: Bool -> T
t2 :: Int -> T
…

You can pattern-match on T1 just fine, but if you try to use it as a
constructor you’ll get “non-bidirectional pattern synonym ‘T1’ used in an
expression”.

On Fri, May 19, 2017 at 1:43 AM, Andrew Butterfield <
Andrew.Butterfield at scss.tcd.ie> wrote:

> All,
>
>  is there any way in Haskell to export a *data* type so that importing
> modules can pattern match,
> but not use the constructors to build anything?
>
> My use case is an AST with invariant - I want the convenience of pattern
> matching
> with the safety of having to build using functions exported by the model
> rather than the constructors directly.
>
>  e.g
>
> given
>
> data T = T1 Bool | T2 Int | TT T T
> t1 :: Bool -> T
> t2 :: Int -> T
> tt :: T -> T -> T
>
> from outside I can write
>
> f(T1 False) =  tt (t1 True) (t2 42)
>
> but not
>
> f(T1 False) = TT (T1 True) (T2 42) ?
>
>
> Regards,
>
> Andrew Butterfield
> School of Computer Science & Statistics
> Trinity College
> Dublin 2, Ireland
>
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