stackoverflow error in following java program

Question

public class testing {    
    testing t=new testing();                

    public static void main(String args[]){   
        testing t1=new testing();  
        t1.fun();  
    }

    void fun(){         
        int a=2;        
        int b=t.fun2(a);  
        System.out.println(a+" "+b);  
    }

    int fun2(int a)  
    {
        a=3;  
        return a;  
    }  
}

why does above code giving following error? I just want to know the reason because it is difficult to expect StackOverFlowError error in this case.

Exception in thread "main" java.lang.StackOverflowError
at com.testing.<init>(testing.java:4)
at com.testing.<init>(testing.java:4)

Ahh, please indent your code. It's much easier to help you when your code is readable. — Jack Satriano
– Jack Satriano, Commented Jun 27, 2012 at 5:32
Also, classes should be capitalized. Look up some code style guidelines. — Jack Satriano
– Jack Satriano, Commented Jun 27, 2012 at 5:34
@jack57 I know coding conventions, just written the code for quick testing. But thanx for advice anyways. — user980089
– user980089, Commented Jun 27, 2012 at 5:39
can someone provide a link where I can refer to understand because I still have some doubts. — user980089
– user980089, Commented Jun 27, 2012 at 6:55

Jigar Joshi · Accepted Answer · 2012-06-27 06:10:51Z

13

You are recursively creating instance of testing

public class testing {    
 testing t=new testing();        
//

}

While creating first instance it will create new Instance by testing t=new testing(); which will again create new instance and so on

edited Jun 27, 2012 at 6:10

answered Jun 27, 2012 at 5:32

Jigar Joshi

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10 Comments

user980089 Over a year ago

I created the object for class testing and according to my understanding it will be created only once. Can you explain how multiple instances are creating?

Jigar Joshi Over a year ago

While creating first instance it will create new Instance by testing t=new testing(); which will again create new instance and so on

Karesh A Over a year ago

when you create the new testing() first time it is creating a object of that class but that class has new testing() and create that object and it will continue..

user980089 Over a year ago

@JigarJoshi Still i don't understant as public class testing { testing t=new testing(); public static void main(String args[]){ System.Out.Print("hello");}} runs successfuly.

yshavit Over a year ago

@user980089 What's happening is that you instantiate a testing object, and the first thing that happens is that this object runs all of its field initializers. In your case, the only initializer is testing t = new testing(), so that runs. Well, what's that do? It instantiates a testing object, which means all of its fields get initialized -- namely, its testing t = new testing(). And so on. But without that first new testing() to get things started, none of them run -- because testing t is an instance variable, not a static variable, and thus only applies to new instances.

|

Kushan · Accepted Answer · 2012-06-27 05:42:36Z

1

try this solution,

public class testing {                

    public static void main(String args[]){   
        testing t1=new testing();  
        t1.fun(t1);  
    }

    void fun(testing t1){         
        int a=2;        
        int b=t1.fun2(a);  
        System.out.println(a+" "+b);  
    }

    int fun2(int a)  
    {
        a=3;  
        return a;  
    }  
}

answered Jun 27, 2012 at 5:42

Kushan

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1 Comment

user980089 Over a year ago

thanx kushan for solution, but i just want to know the reason for error.

RTA · Accepted Answer · 2012-06-27 05:43:15Z

1

You have to create field at class level but instantiate it once where in main

public class Testing {    
    static Testing t1;              

    public static void main(String args[]){   
        t1=new Testing();  
        t1.fun();  
    }

    void fun(){         
        int a=2;        
        int b=t1.fun2(a);  
        System.out.println(a+" "+b);  
    }

    int fun2(int a)  
    {
        a=3;  
        return a;  
    }  
}

answered Jun 27, 2012 at 5:43

RTA

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1 Comment

RTA Over a year ago

some alternate solutions are also there which are able to elude that problem

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