On Thu, 4 May 2006, Jeremy Yallop wrote:
> Thanks for the reply.  That doesn't seem to be what I want, though.  The
> input to k should have type 'f'.  The output should have type 'g'.  Your
> 'k' can be called with values that don't match type 'f':
>
>     # k `C;;
>     - : g = `C
>
> The following does what I want:
>
>     let k (#f as x:f) = (x:g)
>
> I'd like to understand why it behaves differently from the following:
>
>     let k (x:f) = (x:g)

Right, others have shown that you can do what you want with a coercion. The
problem is that you are using exact types ("[" rather than "[>" ) and there
is a mismatch of exact types. ":" is not a coercion, it just tells the type
that should be there. ":>" coerces. x is not of type g, but it can be coerced to
g.

-- Brian