Question

我正在尝试用C ++实现链表的pop功能。我的节点和链接列表类看起来像这样：

//Node class 
template <class T> 
class Node{
    public:
        T data; 
        Node<T> *next;

        Node(T data, Node<T> *next);
        ~Node();
};

template <class T>
Node<T>::Node(T data, Node<T> *next){
    this->next = next;
    this->data = data;
}

template <class T>
Node<T>::~Node(){
    delete this;
}

//LinkedList class
template <class T> 
class  LinkedList{ 
    public:
        //fields
        Node<T> *head;
        int size;

        //methods
        LinkedList();
        void push(T data);
        T pop();   
        void printList();

};

template <class T>
LinkedList<T>::LinkedList(){
    this->head = NULL;
    this->size = 0;
}

template <class T>
void LinkedList<T>::printList(){
    int i = 1;
    while(head){
        std::cout<<i<<": "<<head->data<<std::endl;
        head = head->next;
        i++;
    }
}

int main(){ 
    LinkedList<int> list;

    for (int i = 1; i < 6; ++i)
        list.push(i);

    list.printList();

    for (int j = 0; j < 3; ++j){
        int output=list.pop();
        printf("popped: %d\n", output);
    }

    list.printList();
    return 0;
}

下面是我的pop功能。问题是this-＆gt; head返回NULL。因此我无法更改其值或访问其数据字段。我使用print语句来发现this-＆gt; head返回NULL。我该如何解决这个问题？

template <class T>
T LinkedList<T>::pop(){
    Node<T> *h = this->head;

    //if list is empty
    if (this->size==0){
        return 0;
    }

    //if list has only one node
    if (this->size==1){
        T ret = h->data; 
        this->head = NULL;
        this->size --; 
        return ret; 
    }

    //if list has multiple nodes
    else{
        T ret = this->head->data;
        this -> head = h->next;
        return ret;
    }

    h.~Node<T>();

}

下面是我的推送功能。我已经测试了这个函数并且它工作正常，但如果它没有正确处理节点指针，请告诉我。

template <class T>
void LinkedList<T>::push(T data){
    Node<T> *n = new Node<T>(data, this->head);
    this->head = n;
    this->size++;   
}

Answer 1

template <class T>
Node<T>::~Node(){
    delete this;
}

这是非常非常错误的。你需要完全摆脱它。

更重要的是，printList()正在修改head。这就是调用head时pop()为NULL的原因。使用本地Node*变量迭代列表：

template <class T>
void LinkedList<T>::printList(){
    int i = 1;
    Node<T> *n = head; // <-- here
    while(n){
        std::cout << i << ": " << n->data << std::endl;
        n = n->next;
        i++;
    }
}

此外，pop()没有正确释放节点（如果有的话），并且不总是递减size。它应该看起来更像是这样：

template <class T>
T LinkedList<T>::pop(){
    Node<T> *h = this->head;

    //if list is empty
    if (this->size==0){
        return T(); // <-- not 0! or throw an exception instead...
    }

    //if list has only one node
    if (this->size==1){
        T ret = h->data; 
        this->head = NULL;
        this->size--; 
        delete h; // <-- add this!
        return ret; 
    }

    //if list has multiple nodes

    T ret = this->head->data;
    this->head = h->next;
    this->size--; // <-- add this!

    delete h; // <-- not h.~Node<T>()!

    return ret; // <-- moved down here!
}

或者更简单，这个（不需要单独处理size==1个案例）：

template <class T>
T LinkedList<T>::pop(){
    Node<T> *h = this->head;

    //if list is empty
    if (!h){
        return T();
    }

    //if list has any nodes

    T ret = h->data;
    this->head = h->next;
    this->size--;
    delete h;

    return ret;
}

即使list不为空，C ++ linkedList头也会返回NULL

1 个答案: