help with open vs closed polymorphic variants

help with open vs closed polymorphic variants

[Erick Tryzelaar]

I'm not sure if this is the right term for describing polymorphic
variants, but it seems to me that there are two ways to work with a
complex hierarchy of polymorphic types to get something that's similar
to inheritance. First is to use what I call closed variants, where you
build the tree bottom up like this:

type foo2 = [`Foo2];;
type foo = [foo2 | `Foo];;
type bar = [`Bar];;
type baz = [`Baz];;
type value = [foo|bar|baz];;

The open variants are the other way around:

type value = [`value];;
type foo = [value|`Foo];;
type bar = [value|`Bar];;
type baz = [value|`Baz];;
type foo2 = [foo|`Foo2];;

These both let you call a function on a subset of variants. For closed
variants, you can do:

type 'a t = T of 'a;;
let f (x:[< foo] t) = ();;
f (T (`Foo :> foo));;
f (T (`Foo2 :> foo2));;

but this is a type error: "f (T (`Bar :> bar))". Likewise with open variants:

type 'a t = T of 'a;;
let f (x:[> `Foo] t) = ();;
f (T (`Foo :> foo));;
f (T (`Foo2 :> foo2));;

With "f (T (`Bar :> bar))" being an error as well. However, say we
wanted to allow for a function that works on two subsets of variant
tree. I can do this with closed variants:

let f (x:[< foo|bar] t) = ();;
f (T (`Foo :> foo));;
f (T (`Foo2 :> foo2));;
f (T (`Bar :> bar));;

But now "f (T (`Baz :> baz))" is a type error. However, I can't figure
out if there's an equivalent with open variants. The naive solution
doesn't work:

# let f (x:[< `Foo|`Bar] t) = ();;
# f (make_foo ());;
This expression has type foo t but is here used with type ([< `Foo ] as 'a) t
Type foo = [ `Foo | `value ] is not compatible with type 'a
The second variant type does not allow tag(s) `value

It doesn't even type the first call. Is this impossible?