> Given a polymorphic variant type t with a label `B how does one build
> the type corresponding to t without the label `B.

So that you could write something like:

type t = [ `A | `B | `C ]

let f x =
  match x with
    `B                     -> (* x has type t *)
  | #(t minus [ `B ]) as x -> (* x has type [ `A | `C ] *)

I end up having to write lots of tedious extra types to achieve that
normally... I'd find a subtraction syntax very handy too.

On a similar subject, am I the only person who finds I often need to take a
fixed polymorphic variant value and coerce it so that it can just accept
more constructors e.g. (seriously contrived example)...

type t = [ `A | `B | `C ]

let f (x : t) =
  match x with
    `A -> `D
  | _  -> (x : t :> [> t ])

Something more concise than (x : t :> [> t ]) would be nice: e.g. (x ::> t)
But perhaps that really is a job for camlp4!



David