我正想这样做,但我不相信,如果在移除后指定参考时间仍然有效。
public Box removeBox(int index)
{
Box temp=getBoxes().get(index);//ArrayList.get
getBoxes().remove(index);
return temp;
}
我正想这样做,但我不相信,如果在移除后指定参考时间仍然有效。
public Box removeBox(int index)
{
Box temp=getBoxes().get(index);//ArrayList.get
getBoxes().remove(index);
return temp;
}
这一罚款虽然是<代码>return RecBoxes(index)remove()在单一行文中相同。
Yes it is a valid way of doing it. Java works with values, not references.
<代码>List.remove(> >t t 无效该被删除的物体。
从任何收款中删除任何内容都是无效的。
是的,这一提法仍然有效,因为已经提到了这一提法(<代码>temp)。 页: 1 自动取款机的垃圾收集器在运行时将照顾所有未参保物体。 详情见How Garbage Collection work in Java/a>, 尤其是When an Object变得有资格获得Garbage Collection。
ArrayList#remove
返还移除的元件,如果有的话,也可在
你正在大举, 参考文件有效
举一个例子
页: 1 目的
B. 出席情况 目的
A和B都提到了目标。
现在说,提及B无效
页: 1 目的
西欧和其他国家: 反对(无效)
参考文件A依然没有意义。
Thanks Abhi
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