java.lang.StackOverflowError？

Question

I am sure that, this ques code must be asked in this site. But I am not able to search, This is basic ques, but I am not getting it because of my poor basic concept-

public class A {

A obj = new A();

public static void main(String arg[])
{
    A ob = new A();
}
} 

It is giving java.lang.StackOverflowError ,Why?

Answer 1

每次创建对象A ，都会创建另一个对象A ，该对象也将创建另一个对象A ...

Answer 2

Your class is essentially equivalent to:

public class A {

    A obj;
    public A() {
        obj = new A();
    }  

    public static void main(String arg[]) {
        A ob = new A();
    }
}

Now you see how you got that error? Everytime you create an instance of A , the constructor get's called, which again invokes itself to create another instance, and this goes on filling up the stack till it overflows.

Answer 3

StackOverflow errors occur because there is a very deep recursion within the application. When you instantiate A , you also call the same constructor to create another instance of A and hence, you have a recursive tree and thus causing the stack overflow error.

Hence, the real problem is deep recursive calls to instantiate A .

Answer 4

当创建类型为A的对象时，您正在创建类型为A的新对象，这将创建类型为A的新对象，等等。