递归的流程，java崩溃的原因,该如何处理

递归的流程，java崩溃的原因

Java code

import java.util.*;public class CrashJava{    public String toString()    {        return "CrashJava Address: " + this + "\n";    }    public static void main(String[] args)    {        Vector v = new Vector();        for(int i = 0; i < 10; i++)        {            v.addElement(new CrashJava());        }        System.out.println(v);    }}

[解决办法]
public String toString()
{
return "CrashJava Address: " + this + "\n";/
}

等同于
public String toString()
{
return "CrashJava Address: " + this.toString() + "\n";
}

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