On Sat, Feb 28, 2004 at 10:40:34AM +0100, sejourne kevin wrote:
> A definition for mutable structures look ambiguous to
> me:
> 
> type test = 
> 	{
> 	mutable a:int;
> 	b:int
> 	};;
> 
> let r = {a=0;b=2};;
> 
> let x = r and y = r in 
> assert((x.a==y.a)&&(x.b==y.b)&&(x==y))
> ;;
> 
> Does this always ok ?

Yes, of course. The objects x and y are physically equal (which implies
physical equality of their fields).

However here x == y would fail:

   let x = {a = 0; b = 2}
   let y = {a = 0; b = 2}

while x.b==y.b && x.a==y.a would succeed.

-- 
: Michal Moskal :: http://www.kernel.pl/~malekith :: GCS !tv h e>+++ b++
: When in doubt, use brute force. -- Ken Thompson :: UL++++$ C++ E--- a?

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