Convert type family instances to Int

Question

I have this code:

type family Id obj :: *
type instance Id Box = Int

And I want to make it so I can always get an Int from the Id type family. I recognize that a conversion will be required.

I thought maybe creating a class would work:

class IdToInt a where
  idToInt :: Id a -> Int

instance IdToInt Box where
  idToInt s = s

And that actually compiles. But when I try to use it:

testFunc :: Id a -> Int
testFunc x = idToInt x

I get error:

src/Snowfall/Spatial.hs:29:22:
Couldn't match type `Id a0' with `Id a'
NB: `Id' is a type function, and may not be injective
In the first argument of `idToInt', namely `x'
In the expression: idToInt x
In an equation for `testFunc': testFunc x = idToInt x

So, how can I create a conversion for a type family Id to get an Int?

Based on the answer by ehird, I tried the following but it doesn't work either:

class IdStuff a where
  type Id a :: *
  idToInt :: Id a -> Int

instance IdStuff Box where
  type Id Box = Int
  idToInt s = s

testFunc :: (IdStuff a) => Id a -> Int
testFunc x = idToInt x

It gives error:

src/Snowfall/Spatial.hs:45:22:
Could not deduce (Id a0 ~ Id a)
from the context (IdStuff a)
  bound by the type signature for
             testFunc :: IdStuff a => Id a -> Int
  at src/Snowfall/Spatial.hs:45:1-22
NB: `Id' is a type function, and may not be injective
In the first argument of `idToInt', namely `x'
In the expression: idToInt x
In an equation for `testFunc': testFunc x = idToInt x

Answer 1

You can't. You'll need testFunc :: (IdToInt a) => Id a -> Int. Type families are open, so anyone can declare

type instance Id Blah = ()

at any time, and offer no conversion function. The best thing to do is to put the type family in the class:

class HasId a where
  type Id a
  idToInt :: Id a -> Int

instance IdToInt Box where
  type Id Box = Int
  idToInt s = s

You'll still need the context, though.

Answer 2

You cannot use a function of type IdToInt a => Id a -> Int because there is no way to determine what type a is. The following example demonstrates this.

type family Id a :: *
type instance Id () = Int
type instance Id Char = Int

class IdToInt a where idToInt :: Id a -> Int

instance IdToInt () where idToInt x = x + 1
instance IdToInt Char where idToInt x = x - 1

main = print $ idToInt 1

Because Id () = Id Char = Int, the type of idToInt in the above context is Int -> Int, which is equal to Id () -> Int and Id Char -> Int. Remember that overloaded methods are chosen based on type. Both class instances define idToInt functions that have type Int -> Int, so the type checker cannot decide which one to use.

You should use a data family instead of a type family, and declare newtype instances.

data family Id a :: *
newtype instance Id () = IdUnit Int
newtype instance Id Char = IdChar Int

With a newtype instance, Id () and Id Char are both ints, but they have different types. The type of an Id informs the type checker which overloaded function to use.