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haskellsml

Convert a MOSCOW ML type to HASKELL

I'm trying to convert a SML type to a Haskell type.

type Identifier = string

type 'a Environment = (Identifier * 'a) list
Solution

  • I guess the same code in Haskell could be

    type Identifier = String

type Environment a = [(Identifier, a)]

    Still, I would recommend to use newtypes to increase type safety, like in

    newtype Identifier = Identifier String

newtype Environment a = Environment [(Identifier, a)]