🚀 go-pugleaf

On Sat, 27 Sep 2003 10:59:33 -0400, Gillmer J. Derge (TeamB) wrote:

> "Aleksandar Pecanov" <pecanov@mt.net.mk> wrote in message
> news:pan.2003.09.27.09.12.01.595247@mt.net.mk...
>> The following code terminates my application:
>>
>> public DBValue value( Object val ) {
>>             DBValue v = value(val);
>>             v.setValue(val);
>>
>>             return v;
>>         }
>>
>> It took me quite some time to find out that this was my problem,
>> because I didn't expect the virtual machine to become aware of a
> deadlock.
>> I'm using JDK 1.4.2 (for linux).
>> How does deadlock detection work in Java, and since when was it
> implemented?
> 
> This isn't deadlock.  It's a stack overflow due to infinite recursion.
> You're essentially running out of memory.  value calls value which calls
> value which calls value ... until you eventually run out of space to
> remember the arguments of each method call and a StackOverflowError is
> thrown.

I'm sorry, but it seems I was missunderstood. I know what was suppose to
happen, but no exception was ever throwen. The method resulted in
application shutdown, after it was first invoked. Now, for my
understanding stack overflow needs creation of instances of objects and
methods, which was not the case here.
I'm waiting on some better explanations.