C++超载算子[]

我想为我的deque类实现一个带[]的索引。我为此写了自己的课。。但是，我得到一个错误：在“[”标记之前出现意外的限定符id

也许我不太明白操作符[]是如何工作的，我不能对数组使用常用的[]索引，因为我为我的deque类实现了一个指针结构

我将我的类分为一个.cpp文件和一个.h文件-这里是完整的文件，但可能只有 string&Deque:：operator[]int i{method在.cpp文件中很有趣

.cpp：

.h文件：

#include <iostream>
#include <string>
#include <cstring>

using namespace std;


class Node
{   
private:
    string data="\0";
    Node* next=0;
    Node* prev=0;

public:

Node(){};



Node(string p){
    this->data = p;
}

//KOPIERKONSTRUKTOR MIT POINTER
Node(Node * n) {
    this->data = n->getData();
    this->next = n->getNext();
    this->prev = n->getPrev();

}

//KOPIERKONSTRUKTOR MIT REFERENZPARAMETER
Node(Node& n) {
    Node * tmp = &n;
    this->data = tmp->getData();
    this->next = tmp->getNext();
    this->prev = tmp->getPrev();
    delete tmp;
}

//ZUWEISUNGSOPERATOR(=)
Node& operator=(Node& n) {
    Node * tmp = &n;
    this->data = tmp->getData();
    this->next = tmp->getNext();
    this->prev = tmp->getPrev();
    return *this;
}



string getData(){
    return this->data;
}  

string getData() const{
    return this->data;
}  

void setData(string v) {
    this->data = v;
}

Node * getNext() {
    return this->next;
}

Node * getNext() const{
    return this->next;
}

Node * getPrev() {
    return this->prev;
}  


Node * getPrev() const {
    return this->prev;
}  


void setNext(Node * n) {
    this->next = n;
}

void setPrev(Node * n) {
    this->prev = n;
}    

};



class Deque
{  
private:
    Node* front;
    Node* rear;
    int count;

public:
    Deque()
    {
        front =0 ;
        rear =0;
        count = 0;
    }   

    typedef Node* iterator;


    //ELEMENT AM ANFANG HINZUFUEGEN
    void push_front(string element)
    {
        // Create a new node
        Node* tmp = new Node(element);
        //tmp->data = element;
        //tmp->next = 0;
        //tmp->prev = 0;

        if ( isEmpty() ) {
            // erstes Element hinzufuegen
            front = rear = tmp;
        }
        else {
            // Ganz vorne anhaengen und Pointer umbiegen
            //tmp->next = front;
            tmp->setNext(front);
            //front->prev = tmp;
            front->setPrev(tmp);
            front = tmp;
        }
        count++;
    }

    //ERSTES ELEMENT ENTFERNEN
    string pop_front()
    {
        if ( isEmpty() ) {          
             cout << "Deque ist leer" <<endl;

        }

        //  Wert aus erstem Knoten holen
        string ret = front->getData();

        // Ersten Knoten loeschen und Wert holen
        Node* tmp = front;
        if ( front->getNext() != 0 )
        {
            front = front->getNext();
            front->setPrev(0) ;
        }
        else
        {
            front = 0;
        }
        count--;
        delete tmp;

        return ret;
    }

    //ELEMENT AM ENDE HINZUFUEGEN
    void push_back(string element)
    {          
        // neuen Tmp Knoten erzeugen
        Node* tmp = new Node();
        //tmp->data = element;
        //tmp->next = 0;
        //tmp->prev = 0;

        tmp->setData(element);
        tmp->setNext(0);
        tmp->setPrev(0);

        if ( isEmpty() ) {           
            front = rear = tmp;
        }

        else {
            // hinten an Liste anhaengen, Pointer umbiegen
            //rear->next = tmp;
            rear->setNext(tmp);
            //tmp->prev = rear;
            tmp->setPrev(rear);
            rear = tmp;
        }

        count++;
    }

    void append_Node(Node * firstNode)
    {
        if (isEmpty()) {
            front = rear = firstNode;
        }

        else 
        {
            rear->setNext(firstNode);
            firstNode->setPrev(rear);
        }
    }

    //LETZTES ELEMENT ENTFERNEN
    string pop_back()
    {
        if ( isEmpty() ) {
             throw "Deque ist leer";

        }

        string ret = rear->getData();

        // letzten Knoten loeschen und Pointer umsetzen
        Node* tmp = rear;
        if ( rear->getPrev() != 0 )
        {
            rear = rear->getPrev();
            rear->setNext(0);
        }

        else
        {
            rear = 0;
        }
        count--;
        delete tmp;
        return ret;
    }


    //PEEK ERSTES ELEMENT
    string getFirst()    
    {          
        if ( isEmpty() )
            cout << "Deque ist leer"<<endl;
        return front->getData();
    }

    //PEEK LETZTES ELEMENT
    string getLast()
    {
        if ( isEmpty() )
             cout << "Deque ist leer"<<endl;
        return rear->getData();
    }

    //ANZAHL ELEMENTE IM DECK ZURUECKGEBEN
    int Size()
    {
        return count;
    }

    void setSize(int size){
        this->count = size;
    }

    //DECK LEER?
    bool isEmpty();


    Node * getFront() {
        return front;
    }

    Node * getRear() {
        return rear;
    }

    //"ITERATOR" DEQUE EINMAL DURCHLAUFEN VORNE BIS HINTEN
    //in der ueb3.cpp Datei implementiert
    void ausgeben();

    Deque& operator+=(Deque& a);

    Deque& operator+(Deque& a);

    void iterieren(Deque& d);


    //string& operator[]( int i);
};

---

下面是编译的运算符[]的代码：

string& Deque::operator[](int i) {
        // You need to check whether i is in dequeue range, and, if not in range, throw an exception
    assert( i>=0 && i< this->Size()-1);
    int start = 0;
    string found ;
    if (this->isEmpty()){
                cout<<"Deque ist leer!" <<endl;

    }
    Node * tmp = this->getFront();
    bool gefunden = false;

    while (!gefunden && start < i)
    {
        // In this loop we will always find appropriate element
        if (start!=i) {
        tmp = tmp->getNext();
        if (tmp == 0) {
            cout <<"Es existiert kein " << i<<".tes Element" <<endl;

        }
        start++;
    }  
    if ( start == i) {
        break; // OK, we found one
        //string * found = new String(tmp->getData()); // It won't be compiled
        string * found = new string(tmp->getData());
        cout <<i << ".ter Wert: " <<found<<endl;
        gefunden = true;
        return * found;
    }
    }
        return tmp->getDataRef();
    //return 0; This string is buggy
}

---

我很确定错误消息中包含行号。您应该向我们显示错误来自哪一行。代码太多。请将示例简化为一点大小。使用此代码，您不应该在运算符[]的定义方面遇到任何错误…您的声明和实现都已注释掉。您的运算符[]动态分配它找到的字符串的副本并返回它？这是个坏主意。为什么不返回一个对你找到的字符串的常量引用。至少，如果你想返回一个副本，去掉新的并返回你通过值找到的字符串。+1在Praetorian注释中-实际使用运算符的那一行已经创建了内存泄漏：q[o]；它甚至不会真正编译，因为语法应该是q[o]；这可能就是你的编译器错误。返回的引用从未存储在任何地方，并且会丢失。我看到太多明显的错误，甚至没有尝试编译它。我仍然有一个问题..在cout@user2883596之后，应该有类似于在cout之后抛出std:：runtime_errorEmpty deque的内容

string& Deque::operator[](int i) {
        // You need to check whether i is in dequeue range, and, if not in range, throw an exception
    assert( i>=0 && i< this->Size()-1);
    int start = 0;
    string found ;
    if (this->isEmpty()){
                cout<<"Deque ist leer!" <<endl;

    }
    Node * tmp = this->getFront();
    bool gefunden = false;

    while (!gefunden && start < i)
    {
        // In this loop we will always find appropriate element
        if (start!=i) {
        tmp = tmp->getNext();
        if (tmp == 0) {
            cout <<"Es existiert kein " << i<<".tes Element" <<endl;

        }
        start++;
    }  
    if ( start == i) {
        break; // OK, we found one
        //string * found = new String(tmp->getData()); // It won't be compiled
        string * found = new string(tmp->getData());
        cout <<i << ".ter Wert: " <<found<<endl;
        gefunden = true;
        return * found;
    }
    }
        return tmp->getDataRef();
    //return 0; This string is buggy
}

class Node {
...
std::string& getDataRef() {
        return this->data;
}
...
}