所以我正在制作一个链表，其中我已经成功地实现了插入方法。但接下来我必须将 OrderedLinkedList 实例数据复制到一个新的空实例。问题是当我尝试访问空 LinkedList 的"头"指针时，它说我无法访问头，因为它是常数。以下是错误的完整说明。
完全错误
>------ Build started: Project: betterLinkedList, Configuration: Debug Win32 ------
1> assign3_driver.cpp
1>c:usersw01104311downloadsassign3 (1)assign3orderedlinkedlist.h(67): error C3490: 'head' cannot be modified because it is being accessed through a const object
1> c:usersw01104311downloadsassign3 (1)assign3orderedlinkedlist.h(64) : while compiling class template member function 'void OrderedLinkedList<Type>::copy(const OrderedLinkedList<Type> &)'
1> with
1> [
1> Type=MemberDO
1> ]
1> c:usersw01104311downloadsassign3 (1)assign3orderedlinkedlist.h(60) : see reference to function template instantiation 'void OrderedLinkedList<Type>::copy(const OrderedLinkedList<Type> &)' being compiled
1> with
1> [
1> Type=MemberDO
1> ]
1> c:usersw01104311downloadsassign3 (1)assign3orderedlinkedlist.h(58) : while compiling class template member function 'OrderedLinkedList<Type>::OrderedLinkedList(const OrderedLinkedList<Type> &)'
1> with
1> [
1> Type=MemberDO
1> ]
1> c:usersw01104311downloadsassign3 (1)assign3assign3_driver.cpp(95) : see reference to function template instantiation 'OrderedLinkedList<Type>::OrderedLinkedList(const OrderedLinkedList<Type> &)' being compiled
1> with
1> [
1> Type=MemberDO
1> ]
1> c:usersw01104311downloadsassign3 (1)assign3assign3_driver.cpp(42) : see reference to class template instantiation 'OrderedLinkedList<Type>' being compiled
1> with
1> [
1> Type=MemberDO
1> ]
1>c:usersw01104311downloadsassign3 (1)assign3orderedlinkedlist.h(74): error C2664: 'Node<Type>::Node(const Node<Type> &)' : cannot convert parameter 1 from 'MemberDO' to 'const Node<Type> &'
1> with
1> [
1> Type=MemberDO
1> ]
1> Reason: cannot convert from 'MemberDO' to 'const Node<Type>'
1> with
1> [
1> Type=MemberDO
1> ]
1> No user-defined-conversion operator available that can perform this conversion, or the operator cannot be called
1>c:usersw01104311downloadsassign3 (1)assign3orderedlinkedlist.h(79): error C2664: 'Node<Type>::Node(const Node<Type> &)' : cannot convert parameter 1 from 'MemberDO' to 'const Node<Type> &'
1> with
1> [
1> Type=MemberDO
1> ]
1> Reason: cannot convert from 'MemberDO' to 'const Node<Type>'
1> with
1> [
1> Type=MemberDO
1> ]
1> No user-defined-conversion operator available that can perform this conversion, or the operator cannot be called
1> Generating Code...
1> Compiling...
1> MemberDO.cpp
1> Generating Code...
========== Build: 0 succeeded, 1 failed, 0 up-to-date, 0 skipped ==========
那么我该如何更改这个新的链接列表数据（例如更改头部指针和尾部指针）？这是我的代码：
#ifndef ORDEREDLINKEDLIST_H
#define ORDEREDLINKEDLIST_H
template <class Type>
struct Node
{
Node<Type>();
Type info;
Node*next;
};
template <class Type>
class OrderedLinkedList
{
public:
OrderedLinkedList();
OrderedLinkedList(const OrderedLinkedList& other);
~OrderedLinkedList();
OrderedLinkedList<Type>& operator=(const OrderedLinkedList<Type>& other);
int insert(const Type&);
Type* find(int) const;
Type* get(int) const;
int remove(int);
void clear();
int size() const;
void print() const;
private:
void copy(const OrderedLinkedList<Type>& other);
int length;
Node<Type>* head;
Node<Type>* tail;
};
template <class Type>
Node<Type>::Node(){
next = NULL;
}
template <class Type>
OrderedLinkedList<Type>::OrderedLinkedList()
{
head = NULL;
tail = NULL;
length = 0;
}
template <class Type>
OrderedLinkedList<Type>::OrderedLinkedList(const OrderedLinkedList& other)
{
copy(other);
}
template <class Type>
void OrderedLinkedList<Type>::copy(const OrderedLinkedList& other){
length = other.length;
if( other.head = NULL){
head = NULL;
tail = NULL;
return;
}
head = new Node<Type>(other.head->info);
Node<Type>* current = other.head->next;
Node<Type>* node = head;
while(current!=NULL){
node->next = new Node<Type>(current->info);
if(current->next = NULL){
tail = current;
}
current = current->next;
node = node->next;
}
}
template <class Type>
OrderedLinkedList<Type>& OrderedLinkedList<Type>::operator=(const OrderedLinkedList& other)
{
return *this;
}
template <class Type>
int OrderedLinkedList<Type>::insert(const Type& item)
{
Node<Type>* newNode = new Node<Type>();
Node<Type>* tmp = new Node<Type>();
Node<Type>* prev = new Node<Type>();
newNode->info = item;
bool firstNode = true;
if(head !=NULL){
tmp = head;
while(tmp != NULL && newNode->info.getKey()>tmp->info.getKey()){//put keys in order
if(firstNode){
firstNode = false;
}
prev = tmp;//keeps track of previous node
tmp = tmp->next;
}
if(firstNode){//if first item in the list
head = newNode;
}else{prev->next = newNode;}
if(tmp == NULL){//if last item in the list
newNode->next = NULL;
}else{newNode->next = tmp;}
}else{
head = newNode;
}
while(newNode->next != NULL){
newNode = newNode->next;
}
tail = newNode;
length++;
return newNode->info.getKey();
}
template<class Type>
Type* OrderedLinkedList<Type>::get(int dest) const
{
if(dest>length-1 || dest < 0){
return NULL;
}
Node<Type>*curr = new Node<Type>;
int currCount = 0;
curr = head;
while(currCount < dest){
curr = curr->next;
currCount++;
}
return &(curr->info);//did not delete curr
}
template <class Type>
Type* OrderedLinkedList<Type>::find(int key) const
{
if(key < 0){return NULL;}//checks if negative, if it is return NULL
Node<Type>*curr = new Node<Type>;
int currCount = 0;
curr = head;
while(curr != NULL && curr->info.getKey() != key){
curr = curr->next;
currCount++;
}
if(curr==NULL){
return NULL;
}
return &(curr->info);
}
template <class Type>
int OrderedLinkedList<Type>::remove(int key)
{
return 0;
}
template <class Type>
void OrderedLinkedList<Type>::clear()
{
head = NULL;
tail = NULL;
length = 0;
}
template <class Type>
int OrderedLinkedList<Type>::size() const
{
return length;
}
template <class Type>
void OrderedLinkedList<Type>::print() const
{
}
#endif
这是来自驱动程序源文件的调用
OrderedLinkedList<MemberDO> copy(memberList);
有更多的代码可以保存数据，但我认为在这种情况下不需要它。不过我可能是错的，所以如果您需要更多代码，请告诉我。提前感谢您的帮助！！
您的Node类模板没有采用Type（其模板参数）的构造函数，但您反复将其视为：
例如：
head = new Node<Type>(other.head->info);
和：
node->next = new Node<Type>(current->info);
唯一的单个参数构造函数是实现定义的复制和移动构造函数，它将采用Node<Type>而不是Type（或严格意义上对Node<Type>的某种形式的引用）。