我无法理解在c ++中使用嵌套类的概念。
我正在尝试为链表创建一个迭代器。我已将它声明为LinkedList.h的私有范围内的嵌套类
但是我在哪里定义类Iterator的功能?现在,我试图在文件的底部以及所有LinkedList的功能中这样做。但无论我如何扭曲和旋转范围定义,我都无法使其发挥作用。我应该以不同的方式做这件事吗?
Iterator类
class Iterator {
private:
LinkedList<T>* l;
Node<T>* node;
public:
Iterator(LinkedList<T> *ll, Node<T> *n);
T get();
int hasNext();
void moveNext();
};
四个函数定义
template <typename T>
typename LinkedList<T>::Iterator<T>::Iterator(LinkedList<T> *ll, Node<T> *n) {
l = ll;
node = n;
}
template <typename T>
T LinkedList<T>::Iterator<T>::get() {
return node->getData();
}
template <typename T>
int LinkedList<T>::Iterator<T>::hasNext() {
return node->getNext();
}
template <typename T>
void LinkedList<T>::Iterator<T>::moveNext() {
node = node->getNext();
}
我得到的错误是:
&#34;使用未定义的类型:LinkedList :: Iterator&#34;
&#34;语法错误:意外类型LinkedList&#34;
&#34;语法错误:缺少&#39;;&#39;之前&#39; {&#39;&#34;
还有更多。
我非常非常失落。
编辑:
LinkedList.h文件的完整内容:
#ifndef LINKEDLIST_H
#define LINKEDLIST_H
#include "Node.h"
#include "List.h"
#include <iostream>
template <typename T>
class LinkedList : public List {
private:
int n;
Node<T> *first;
Node<T> *last;
template <typename T>
class Iterator {
private:
LinkedList<T>* l;
Node<T>* node;
public:
Iterator(LinkedList<T> *ll, Node<T> *n);
T get();
int hasNext();
void moveNext();
};
public:
LinkedList();
LinkedList(const LinkedList & ll);
~LinkedList();
int size();
void clear();
void addFirst(T data);
void addLast(T data);
T removeFirst();
T removeLast();
T getFirst();
T getLast();
Node<T>* getFirstNode() const;
void addAt(int pos, T data);
T removeAt(int pos);
T getAt(int pos);
LinkedList& operator=(const LinkedList<T> &right);
T operator[](int i);
LinkedList& operator+(const LinkedList<T> &right);
LinkedList& operator+=(const LinkedList<T> &right);
friend std::ostream& operator<<(std::ostream &os, const LinkedList<T> & ll);
typename LinkedList<T>::Iterator<T> getIterator();
void mergeSort();
};
template <typename T>
LinkedList<T>::LinkedList() {
this->n = 0;
this->first = 0;
this->last = 0;
}
template <typename T>
LinkedList<T>::LinkedList(const LinkedList & ll) {
this-> n = 0;
this-> first = 0;
this-> last = 0;
Node *temp = ll.first;
while(temp) {
addLast(temp->getData());
temp = temp->getNext();
}
}
template <typename T>
void LinkedList<T>::addFirst(T data) {
Node<T> *p = new Node<T>(data, first);
first = p;
if(!n)
last = p;
n++;
}
template <typename T>
void LinkedList<T>::addLast(T data) {
Node<T> *p = new Node<T>(data, 0);
if(!n)
first = last = p;
else {
last->next = p;
last = p;
}
n++;
}
template <typename T>
T LinkedList<T>::removeFirst() {
T a = 0;
if(!n)
throw "Can't retrieve element from empty list!";
a = first->getData();
Node<T> *p = first->next;
delete first;
first = p;
n--;
return a;
}
template <typename T>
T LinkedList<T>::removeLast() {
T a = 0;
if(!n)
throw "Can't retrieve element from empty list!";
if(n == 1) {
a = last->getData();
delete first;
first = last = 0;
}
else {
a = last->getData();
Node<T> *p = first;
while(p->next->next != 0)
p = p->next;
delete p->next;
p->next = 0;
last = p;
}
n--;
return a;
}
template <typename T>
T LinkedList<T>::getFirst() {
if(n < 1)
throw "Can't retrieve element from empty list!";
return first->getData();
}
template <typename T>
T LinkedList<T>::getLast() {
if(n < 1)
throw "Can't retrieve element from empty list!";
return last->getData();
}
template <typename T>
Node<T>* LinkedList<T>::getFirstNode() const {
return first;
}
template <typename T>
int LinkedList<T>::size() {
return n;
}
template <typename T>
T LinkedList<T>::getAt(int pos) {
if(pos >= n)
throw "Element index out of bounds!";
Node<T> *temp = first;
while(pos > 0) {
temp = temp->next;
pos--;
}
return temp->getData();
}
template <typename T>
void LinkedList<T>::clear() {
Node<T> *current = first;
while(current) {
Node<T> *next = current->next;
delete current;
if(next)
current = next;
else
current = 0;
}
}
template <typename T>
void LinkedList<T>::addAt(int pos, T data) {
if(pos >= n)
throw "Element index out of bounds!";
if(pos == 0)
addFirst(data);
else {
Node<T> *temp = first;
while(pos > 1) {
temp = temp->next;
pos--;
}
Node<T> *p = new Node<T>(data, temp->next);
temp-> next = p;
n++;
}
}
template <typename T>
T LinkedList<T>::removeAt(int pos) {
if(pos >= n)
throw "Element index out of bounds!";
if(pos == 0)
return removeFirst();
if(pos == n - 1)
return removeLast();
else {
Node<T> *p = first;
while(pos > 1) {
p = p->next;
pos--;
}
T a = p->next->getData();
Node<T> *temp = p->next;
p->next = p->next->next;
delete temp;
n--;
return a;
}
}
template <typename T>
LinkedList<T>::~LinkedList() {
clear();
}
template <typename T>
LinkedList<T>& LinkedList<T>::operator=(const LinkedList<T> &right) {
if(this != &right) {
n = 0;
first = 0;
last = 0;
Node<T> *temp = right.first;
while(temp) {
addLast(temp->getData());
temp = temp->getNext();
}
}
return *this;
}
template <typename T>
T LinkedList<T>::operator[](int i) {
return getAt(i);
}
template <typename T>
LinkedList<T>& LinkedList<T>::operator+(const LinkedList<T> &right) {
Node<T> *temp = right.first;
while(temp) {
addLast(temp->getData());
temp = temp->getNext();
}
return *this;
}
template <typename T>
LinkedList<T>& LinkedList<T>::operator+=(const LinkedList<T> &right) {
Node<T> *temp = right.first;
while(temp) {
addLast(temp->getData());
temp = temp->getNext();
}
return *this;
}
template <typename T>
std::ostream& operator<<(std::ostream &os, const LinkedList<T> &ll) {
Node<T> *temp = ll.getFirstNode();
while(temp) {
os<<temp->getData()<<std::endl;
temp = temp->getNext();
}
return os;
}
template <typename T>
typename LinkedList<T>::Iterator<T>::Iterator(LinkedList<T> *ll, Node<T> *n) {
l = ll;
node = n;
}
template <typename T>
T LinkedList<T>::Iterator<T>::get() {
return node->getData();
}
template <typename T>
int LinkedList<T>::Iterator<T>::hasNext() {
return node->getNext();
}
template <typename T>
void LinkedList<T>::Iterator<T>::moveNext() {
node = node->getNext();
}
template <typename T>
typename LinkedList<T>::Iterator<T> LinkedList<T>::getIterator() {
return new Iterator<T>(*this, *first);
}
#endif
答案 0 :(得分:2)
您需要使用LinkedList类的模板类型名称(这是有意义的)。这意味着您不需要Iterator类的单独模板类型名称。
template <typename T>
class Iterator {
应该只是
class Iterator {
然后成员将被定义为
template <typename T>
T LinkedList<T>::Iterator::get() {
OR 。然后,至少您必须更改T
的名称,因为它会影响LinkedList
的名称。像
template <typename V>
class Iterator {
...
V get();
...
memeber声明将采用
形式template <typename T>
template <typename V>
V LinkedList<T>::Iterator<V>::get() {
答案 1 :(得分:1)
我认为这样做会:
template <typename T>
LinkedList<T>::Iterator::Iterator(LinkedList<T> *ll, Node<T> *n) {
l = ll;
node = n;
}
template <typename T>
T LinkedList<T>::Iterator::get() {
return node->getData();
}
template <typename T>
int LinkedList<T>::Iterator::hasNext() {
return node->getNext();
}
template <typename T>
void LinkedList<T>::Iterator::moveNext()
{
node = node->getNext();
}