Question

当我尝试重载复制构造函数时，它给出了以下错误代码当我试图重载＆lt;＆lt;时，我遇到了类似的错误消息操作

我通过定义内部标题

来修复它

然而，这次同样的伎俩不起作用..

如何修复此问题以及导致此错误消息的原因是什么？

这是我的复制构造函数定义

template <class ItemType>
void List342<ItemType>::operator=(List342 &source)
{
Node *sNode, *dNode, *insNode;

if (this == &source) return;

this->Clear();
if (source.head == NULL)
{
    return;
}
dNode = new Node;
dNode->value = (source.head)->value;
this->head = dNode;
sNode = (source.head)->next;
while (sNode != NULL)
{
    insNode = new Node;
    insNode->value = sNode->value;
    dNode->next = insNode;
    dNode = dNode->next;
    sNode = sNode->next;
}
}

这是我的头文件。

#ifndef LIST342_H_
#define LIST342_H_

#include <iostream>
#include <string>
using namespace std;

template <class ItemType>
class List342 {

friend ostream& operator<<(ostream &outStream, const List342 &list)
{
    Node *pNode;

    pNode = list.head;
    while (pNode != NULL)
    {
        outStream << *pNode->data;
        pNode = pNode->next;
    }
    return outStream;
}

public:
List342();
List342(List342<ItemType> &);
~List342();

bool BuildList(string FileName);
bool Insert(ItemType *obj);
bool Remove(ItemType target, ItemType &result);//check
bool Peek(ItemType target, ItemType &result);//check
bool isEmpty();
void ClearList();
bool Merge(List342 &list1, List342 &list2);

List342 operator+(const List342 &list) const;
List342 operator+=(const List342 &list);

bool operator==(const List342 &list) const;
bool operator!=(const List342 &list) const;

void operator=(List342 &);

private:
struct Node {
    ItemType *data;
    Node *next;
};
Node *head;
};
#endif /* LIST342_H_ */

复制赋值运算符重载类模板

0 个答案: