On Fri, Feb 27, 2004 at 02:29:50PM -0600, Kevin S. Millikin wrote:
> I'm trying to figure out what I can rely on about physical equality. 
>  I've checked the OCaml manual but can't seem to find what I want to 
> know.
> 
> Presume
> 
> # type t = V0 | V1 of int;;
> type t = V0 | V1 of int
> 
> # V0 == V0;;
> - : bool = true
> 
> V0's are the same.  Is this guaranteed?

Yes.

> 
> # V1(0) == V1(0);;
> - : bool = false
> 
> V1's are different.  Is this guaranteed?

What do you mean by guaranteed? This is true, but it might become
false:

a) because the compiler optimize a common constant subexpression

b) because the runtime might make these 2 values become the same (ie
   by hash-consing).

c) there is a tricky issue regarding floating point compares for
NaN. Read the source code or some previous posting on this (I never
used NaN since I am not a numerical coder).

As far as I know, neither a nor b is currently true (both in
Ocaml3.07pl2 and in the latest CVS), but I could imagine that some
later version of ocaml might do a or b (even if I believe this
unlikely).

The only reasonable implication about physical equality is 
  a == b  implies a = b

(ie physical equality implies structural equality)


Why would you want a guarantee that V1 0 is not physically equal to V1
0? I tend to think that making such an hypothesis is dangerous and
wrong, even if the current implementation demonstrate it.


Regards.

-- 
Basile STARYNKEVITCH         http://starynkevitch.net/Basile/ 
email: basile<at>starynkevitch<dot>net 
aliases: basile<at>tunes<dot>org = bstarynk<at>nerim<dot>net
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