RE: [Sml-implementers] Could raise be non-expansive?

[Robert H. <Rob...@cs...>]

I intend the second interpretation.  The explicit raise is indistinguishable
from the implicit raise caused by matching an expression against a refutable
pattern.

The issue has nothing whatsoever to do with "generating names", whatever
those are.

Bob

-----Original Message-----
From: Dave Berry [mailto:da...@ta...]
Sent: Friday, September 07, 2001 6:50 PM
To: Robert Harper; 'Dave Berry'; sml...@li...
Subject: RE: [Sml-implementers] Could raise be non-expansive?


At 11:54 07/09/2001, Robert Harper wrote:
>I mistakenly reversed the senses of "expansive" and "non-expansive", as
must
>be obvious to everyone.  Sorry about that.


Hi Bob,

I'm afraid I'm now confused, as I can see two possible interpretations of
your message.  Please can you tell me which you intended?

Example:
	exception Foo
	val it = raise Foo
raise Foo has type 'a; can 'a be generalised?

First interpretation:  Yes, 'a can be generalised.  The definition should
be changed to allow this, thus removing an inconsistency between an
explicit raise and the implicit raise caused by a refutable pattern.

Second interpretation:  No, 'a can not be generalised.  The definition
should be changed to also ban the generalisation of type variables in any
expression matched against a refutable pattern, regardless of whether the
rhs expression is expansive.

FWIW, the Commentary only discusses the question of generating exception
names.  If I understand correctly, simply raising an exception doesn't
generate a new name.

Dave.