Re: Weird entry in the exception table for monitorenter/monitorexit

On Wed, 18 Jul 2007 05:16:46 -0700, Douwe <dmvos2000@[EMAIL PROTECTED]
> wrote,
quoted or indirectly quoted someone who said :

>result (I've removed the parts that are irrelevant)
>
>  public void test();
>     0  ldc <Class TestClass> [1]
>     2  dup
>     3  astore_1
>     4  monitorenter
>     5  aload_0 [this]
>     6  iconst_1
>     7  putfield TestClass.a : int [12]
>    10  aload_1
>    11  monitorexit
>    12  goto 18
>    15  aload_1
>    16  monitorexit
>    17  athrow
>    18  return
>      Exception Table:
>        [pc: 5, pc: 12] -> 15 when : any
>        [pc: 15, pc: 17] -> 15 when : any
>
>I do understand that after the monitor is entered it is very im****tant
>that it leaves it again with the instruction monitorexit. In normal
>case where no exceptions occur it will call monitorexit at the address
>11. But if an exception occurs it has to make sure monitorexit is
>called and therefore the exception table defines in the first entry
>that on any exception the VM jumps to address 15. Until here
>everything is understandable, but what is the second entry in the
>exception table telling me? if an exception occurs betweern 15 and 17
>then jump to 15 ... doesn't this cause an endless loop?

I will make a guess that the implementation of exception propagation
checks for a "in monitor" bit and deals with it.
-- 
Roedy Green Canadian Mind Products
The Java Glossary
http://mindprod.com