Implementing DataType abstraction in Haskell

Question

I want to implement am Abstract Datype in haskell. Given a moule with a defined type, say Mytype:

module A (myType,MyType) where

type MyType = Float

mytype :: Float -> MyType
myType f = f

Implemented internally as a single Float, I exported only a function to construct a value given a Float, and the type itself. The problem is, when I load that module, I can acces the implementation.

Given:

module B where
import A

data OtherType = One MyType
               | Two MyType MyType
               deriving Show

I can construct an object of type OtherType like this:

One $ mytype 1.0

Or like this:

One $ (1.0 :: Float)

With a real abstraction I shouldn't be able to do that!

How can I export the type Mytype, in a way such that I can only construct values from my constructor functions

type doesn't create anything new. It just creates a convenient name for an already existing type. Once the code is compiled, your type synonyms are gone, and mytype will have type Float -> Float — user2407038
– user2407038, Commented Feb 15, 2014 at 4:50
type behaves equivalent to typedef in C if you're familiar with that. You're looking for data or newtype. — Cubic
– Cubic, Commented Feb 15, 2014 at 13:30

lando · Accepted Answer · 2014-02-15 01:36:21Z

2

You can create an Algebraic Datatype instead:

module A (myType,MyType) where

data MyType = MyType Float

mytype :: Float -> MyType
myType f = MyType f

Then, trying to evaluate things like

One (MyType 3.0)

throws "Not in scope: data constructor `MyType'"

answered Feb 15, 2014 at 1:36

lando

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1 Comment

Carl Over a year ago

Or newtype. The key point is to not use type, because that's not what it does.

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