brogoff wrote:
> On Thu, 4 May 2006, Luc Maranget wrote:
> 
>>I cannot really explain why it matters, but I can supply a minimal (?) example
>>
>>type f = [`A ]
>>type g = [f | `C]
>>
>>let k (x:f) = (x:g);;
>>               ^
>>This expression has type f but is here used with type g
>>The first variant type does not allow tag(s) `C
> 
> 
> Not enough polymorphism, the error message seems clear
> 
> type 'a h = 'a constraint 'a = [> `A];;
> let k (x : 'a h) = (x : g)

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)

Jeremy.