关于容器的一个有趣话题,该怎么处理

关于容器的一个有趣话题
List list = new ArrayList();
List vector= new Vector();
 
list.add(1);
list.add(2);
vector.add(3);
vector.add(4);
list.add(vector);
System.out.println(list);
vector.addAll(); //??为什么不能用add（）方法
System.out.println(vector);
//输出结果很有意思，但我不明白为什么
//最后一个输出结果如下：1，2，3，4，this connection()

//怎么会有this connection() ？？？？？？？？？？？？？？？？



[解决办法]
是this Collection不是this connection

list = [1,2];
vector=[3,4];
list.add(vector);//将vector加入list
list=[1, 2, [3, 4]]


vector.addAll(list);//list里面已经包含了vector，此时再将list添加到vector，可以简单理解为死循环了。
vector=[3, 4, 1, 2, (this Collection)]=[3, 4, 1, 2, [3, 4, 1, 2, [3, 4, 1, 2, (this Collection)]]]
其实this Collection就是vector
[解决办法]
2楼的哥们已经从逻辑说明原因了，具体的代码可以看下面的代码
Vector的toString方法是super.toString()，而具体执行的方法

Java code

/**     * Returns a string representation of this collection.  The string     * representation consists of a list of the collection's elements in the     * order they are returned by its iterator, enclosed in square brackets     * (<tt>"[]"</tt>).  Adjacent elements are separated by the characters     * <tt>", "</tt> (comma and space).  Elements are converted to strings as     * by {@link String#valueOf(Object)}.     *     * @return a string representation of this collection     */    public String toString() {        Iterator<E> i = iterator();    if (! i.hasNext())        return "[]";    StringBuilder sb = new StringBuilder();    sb.append('[');    for (;;) {        E e = i.next();        sb.append(e == this ? "(this Collection)" : e);        if (! i.hasNext())        return sb.append(']').toString();        sb.append(", ");    }    }