Just a quick hacked-together test:

public class test {
  public static void main(String[] args) {
    m1();
  }
  private static void m1() {
    m2();
  }
  private static void m2() {
    Throwable t = new Throwable();
    t.fillInStackTrace();
    StackTraceElement[] ste = t.getStackTrace();
    for (int i = 0; i < ste.length; i++) {
      System.out.println(i + " - " + ste[i]);
    }
  }
}

I think that will do the trick.

-- 
Frank W. Zammetti
Founder and Chief Software Architect
Omnytex Technologies
http://www.omnytex.com
AIM: fzammetti
Yahoo: fzammetti
MSN: [EMAIL PROTECTED]

On Thu, February 16, 2006 3:37 pm, Frank W. Zammetti said:
> Hey yeah, I think your right!  It *still* may be a little heavy, but one
> would definitely think less so.  Off to play for a few minutes...
>
> --
> Frank W. Zammetti
> Founder and Chief Software Architect
> Omnytex Technologies
> http://www.omnytex.com
> AIM: fzammetti
> Yahoo: fzammetti
> MSN: [EMAIL PROTECTED]
>
> On Thu, February 16, 2006 3:29 pm, [EMAIL PROTECTED] said:
>> Michael Jouravlev replied
>>> On 2/16/06, Frank W. Zammetti <[EMAIL PROTECTED]> wrote:
>>> > In the method you want to "protect", immediately throw an exception
>>> > and catch it.  Then, parse the stack trace and see who the
>>> caller was.
>>> > If it's not a class you want to have access to the method, throw an
>>> > IllegalAccessException.
>>>
>>> I guess it can be costly if the callers are mostly the ones
>>> that you want to allow access to. On the other hand, I don't
>>> know the metrics on how costly is to create and throw an exception.
>>
>> I don't think you'd have to throw and catch it.  I think it's sufficient
>> to create an instance of Throwable to check the stack trace.
>>
>>  - George
>>    http://www.idiacomputing.com
>>
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>
>


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