我的代码存在问题，我将尝试将其简化为概念性问题。
我有一个名为Declare @YourTable table (mis int,mnth varchar(25),IPTV decimal(18,9))
Insert Into @YourTable values (0,'January',0.2026856),(0,'February',0.4860267),(0,'March',0.2650674),(0,'April', 0.1904701),(1,'January',0.266809),(1,'February',0.310241),(1,'March', 0.2083876),(1,'April', 0.3401039),(2,'January',0.307787),(2,'February',0.5276488),(2,'March',0.3037852),(2,'April',0.1822988),(3,'January',0.8107423),(3,'February',0.2430134),(3,'March',0.1988006),(3,'April',0.2539602)
--below is the solution
select *,sum(iptv)over(partition by mnth order by mis) as Wanted
from @yourtable
的班级，它是PunctuationOccurrenceList的{{1}}。 LinkedList只是一个没有任何指针的数据类。
PunctuationOccurrence的析构函数如下：
PunctuationOccurrence LinkedList有template <class Object>
LinkedList<Object>::~LinkedList()
{
/* Here, you could just delete the headerNode, since everything else is linked to it (the point of a LinkedList!!), and will
* also be deleted!! */
if (this->headerNode)
delete this->headerNode;
this->lastNode = 0;
}
个具有析构函数的内容：
LinkedList我必须为ListNode编写复制构造函数。但是，当template <class Object>
ListNode<Object>::~ListNode()
{
if (this->nextNode)
delete this->nextNode;
}
超出范围时，事情会变得奇怪。在测试PunctuationOccurrenceList模块时，它可以正常工作。但是，当用作另一个类的对象成员时，如果它不为空，则会发生分段错误。
我知道rule of three说我应该写复制构造函数。我该怎么写呢？
答案 0 :(得分：2)
LinkedList的复制构造函数应该创建整个列表的副本;它应该复制原始列表中的每个节点。