From: "Erick Tryzelaar" <idadesub@users.sourceforge.net>
To: "OCaml List" <caml-list@yquem.inria.fr>
Subject: help with open vs closed polymorphic variants
Date: Sat, 15 Mar 2008 21:15:09 -0700 [thread overview]
Message-ID: <1ef034530803152115n27d005d4kf7a136309bc90f9d@mail.gmail.com> (raw)
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?
reply other threads:[~2008-03-16 4:15 UTC|newest]
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