Why does the java code below give StackOverflow error?

Question

I am getting StackOverflow error while executing the below:

public class StackOverflow7 {
    StackOverflow7 obj = new StackOverflow7();
    int finalCount = 0;
    public static void main(String[] args) {
        for(int i = 1 ; i <= 5 ; i++)
        System.out.println(i);

        StackOverflow7 localObj = new StackOverflow7();
        localObj.count(88);
        System.out.println("Final Count :: " + localObj.finalCount);
    }

    private void count(int num){
        finalCount = finalCount + num;
    }
}

dryairship · Accepted Answer · 2016-07-08 12:18:03Z

4

This line:

StackOverflow7 obj = new StackOverflow7();

is always called when you create an object of StackOverflow7, which you are doing in this line itself. Thus, this line recursively calls itself, until you get a StackOverflow error.

answered Jul 8, 2016 at 12:18

dryairship

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Comments

Thomas · Accepted Answer · 2016-07-08 12:18:03Z

1

You get the stack overflow because of this line: StackOverflow7 obj = new StackOverflow7();. Whenever you create a new instance, it is called and thus you create a new instance and create a new instance and so on. In your stack trace you should thus see a lot of <clinit> lines.

You start the whole thing by calling StackOverflow7 localObj = new StackOverflow7(); in your main method.

To fix this either make obj a static field or remove it altogether, since you're not using it anyways.

answered Jul 8, 2016 at 12:18

Thomas

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Comments

Jens · Accepted Answer · 2016-07-08 12:18:28Z

0

Because of this line:

StackOverflow7 obj = new StackOverflow7();

Every time you create a new object this line will executet and try to create an other object

answered Jul 8, 2016 at 12:18

Jens

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