Deriving a type and its dependencies

Question

I've been playing with newtype wrappers for my indexes to avoid bugs, and I have some code like this:

{-# LANGUAGE GeneralizedNewtypeDeriving #-}

newtype PersonIdx = PersonIdx Int
  deriving (Enum, Eq, Integral, Num, Ord, Real, Show)

To derive Integral, one must derive all its dependencies as well (hence the above list of type-classes).

Is there a way to make it derive all the dependencies of Integral along with it? I would imagine something like:

newtype PersonIdx = PersonIdx Int
  deriving (Integral(..))

score 6 · Accepted Answer · answered Jan 18 '10 at 17:37

6

No, there is no shortcut. You'll have to derive from all dependent typeclasses explicitly, but can easily find these by typing :i Integral in ghci.

answered Jan 18 '10 at 17:37

Dario

2

I wonder if there's a principled reason for this, or simply that no one has implemented the shortcut in question. – Phil Miller Jan 18 '10 at 18:03
It is somewhat in the philosophy of Haskell that everything be explicit. There are very few "shortcuts" in the language, and this example is no exception. – Martijn Jan 19 '10 at 10:15

1 Answers1