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Join Date: Nov 2009

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Reputation: abalasuriya is an unknown quantity at this point

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abalasuriya abalasuriya is offline

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Java

23 Days Ago

 Help with Code Tags 
 Java Syntax (Toggle Plain Text) 
public class Test {
  static Test t = new Test();
      public static void main(String[] args) {
     Test a = new Test();
       System.out.println(a.t);
    }}
public class Test {
  static Test t = new Test();
      public static void main(String[] args) {
     Test a = new Test();
       System.out.println(a.t);
    }}

This results stack overflow why

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Join Date: Feb 2006

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masijade

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Nearly a Posting Maven

23 Days Ago

No it doesn't.

Does the real class maybe have a constructor?
And does that constructor maybe also call new on itself?
If so, what do think will happen if every call to new results in another to call new?

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Join Date: Nov 2009

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Reputation: abalasuriya is an unknown quantity at this point

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abalasuriya abalasuriya is offline

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23 Days Ago

 Help with Code Tags 
 Java Syntax (Toggle Plain Text) 
public class Test {
   Test t = new Test();
      public static void main(String[] args) {
     Test a = new Test();
       System.out.println(a.t);
    }}
public class Test {
   Test t = new Test();
      public static void main(String[] args) {
     Test a = new Test();
       System.out.println(a.t);
    }}

Sorry!! I made a mistake code should be correct as above and then stack overflow results why ?

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Join Date: Feb 2006

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masijade

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Nearly a Posting Maven

23 Days Ago

Because without the static, that Test t = new Test(); is an instance variable, that is instantiated with every call to new. So, you call new Test() in main, which triggers another new Test, which triggers another new Test, which triggers another new Test, which triggers another new Test, which trig.....

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Join Date: Jun 2008

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Reputation: shobhit123 is an unknown quantity at this point

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shobhit123 shobhit123 is offline

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22 Days Ago

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Originally Posted by abalasuriya

Help with Code Tags

Java Syntax (Toggle Plain Text)
public class Test {
  static Test t = new Test();
      public static void main(String[] args) {
     Test a = new Test();
       System.out.println(a.t);
    }}
public class Test { static Test t = new Test(); public static void main(String[] args) { Test a = new Test(); System.out.println(a.t); }}
This results stack overflow why

The code does not give any stack overflow error.
it runs fine.

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Join Date: Feb 2006

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masijade

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Nearly a Posting Maven

-1

22 Days Ago

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Originally Posted by shobhit123

The code does not give any stack overflow error.
it runs fine.

Which has already been covered. Did you read the thread or just knee jerk reply without bothering to find out what has gone before?

Java Programmer and Sun Systems Administrator

----------------------------------------------

Debugging is twice as hard as writing the code in the first place. Therefore, if you write the code as cleverly as possible, you are, by definition, not smart enough to debug it.
--Brian Kernighan

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Join Date: Jun 2008

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Reputation: shobhit123 is an unknown quantity at this point

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shobhit123 shobhit123 is offline

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-4

22 Days Ago

•

Originally Posted by masijade

Which has already been covered. Did you read the thread or just knee jerk reply without bothering to find out what has gone before?

Yes i didnt see that properly.My mistake.
some people do think themselves smart using slang languages, you are indeed one of them.
anyways..

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peter_budo peter_budo is online now

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Code tags enforcer

21 Days Ago

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Originally Posted by shobhit123

Yes i didnt see that properly.My mistake.
some people do think themselves smart using slang languages, you are indeed one of them.
anyways..

I do recommend you take time to read other responses as often people will reply to something that was already corrected and then wonder why they are hatted by masses. Flaming other and calling him "smart mouth" because he highlighted your mistake is bad idea, you would better do just saying "I'm sorry, my bad".
If you ignore previously provided feedback on this forum or real life you are more likely to fail on next step...

Last edited by peter_budo; 21 Days Ago at 5:42 am.

Learn to see in another's calamity the ills which you should avoid.
Publilius Syrus
(~100 BC)
LJC - London Java Community, Graduate & Undergraduate Software Development Community, JAVAWUG (Java Web User Group), The London Android Group

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Join Date: Jun 2008

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Reputation: shobhit123 is an unknown quantity at this point

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shobhit123 shobhit123 is offline

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21 Days Ago

•

Originally Posted by peter_budo

I do recommend you take time to read other responses as often people will reply to something that was already corrected and then wonder why they are hatted by masses. Flaming other and calling him "smart mouth" because he highlighted your mistake is bad idea, you would better do just saying "I'm sorry, my bad".
If you ignore previously provided feedback on this forum or real life you are more likely to fail on next step...

I do agree my mistake, but there is a way to point out the mistakes too,not simply saying something.We are all professionals.Throwing stones like this should not be done.
I accepted my mistake.
Anyways, lets not drag this issue more and carry on with some technical disscussion.

Thanks,
Shobhit

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Join Date: Mar 2008

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Reputation: MoZo1 is an unknown quantity at this point

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MoZo1

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sollution

#10

21 Days Ago

This is a sollution for the 2nd question where the static has been taken out:

 Help with Code Tags 
 JAVA Syntax (Toggle Plain Text) 
public class Test {
 
    Test t = new Test();
 
    public static void main(String[] args) {
        try {
            Test a = new Test();
            System.out.println("meowmeow");
            System.out.println(a.t);
        } catch (Throwable e) {
            System.out.println("Error:");
            System.out.println(e.getClass());
            System.out.println(e.getMessage());
            System.out.println(e.getCause());
            e.printStackTrace();
        }
    }
}
public class Test {

    Test t = new Test();

    public static void main(String[] args) {
        try {
            Test a = new Test();
            System.out.println("meowmeow");
            System.out.println(a.t);
        } catch (Throwable e) {
            System.out.println("Error:");
            System.out.println(e.getClass());
            System.out.println(e.getMessage());
            System.out.println(e.getCause());
            e.printStackTrace();
        }
    }
}

So when you get stack overflow, you just have to look at the stack. The above code does this, and reports the error in a more convenient way. Also notice that I've cut apart that 2 line of code with a meowing, so now you can know where the error is coming from (before or after).
So now it's trivial, that when you created a new Test, it created a new Test for itself, which also created a new Test for itself, which...

Other fix than removing the redundancy altogether:

 Help with Code Tags 
 JAVA Syntax (Toggle Plain Text) 
public class Test {
 
    Test t = null;
 
    public Test getTest() {
        if (this.t == null) {
            this.t = new Test();
        }
        return this.t;
    }
 
    public static void main(String[] args) {
        try {
            Test a = new Test();
            System.out.println("meowmeow");
            System.out.println(a);
            System.out.println(a.getTest());
            System.out.println(a.getTest().getTest());
            System.out.println(a.getTest().getTest().getTest());
        } catch (Throwable e) {
            System.out.println("Error:");
            System.out.println(e.getClass());
            System.out.println(e.getMessage());
            System.out.println(e.getCause());
            e.printStackTrace();
        }
    }
}
public class Test {

    Test t = null;

    public Test getTest() {
        if (this.t == null) {
            this.t = new Test();
        }
        return this.t;
    }

    public static void main(String[] args) {
        try {
            Test a = new Test();
            System.out.println("meowmeow");
            System.out.println(a);
            System.out.println(a.getTest());
            System.out.println(a.getTest().getTest());
            System.out.println(a.getTest().getTest().getTest());
        } catch (Throwable e) {
            System.out.println("Error:");
            System.out.println(e.getClass());
            System.out.println(e.getMessage());
            System.out.println(e.getCause());
            e.printStackTrace();
        }
    }
}

Lazy evaluation!