使用双向链表的队列中的 Remove() 函数答案

【问题标题】：Remove() function in Queue using a doubly linked list使用双向链表的队列中的 Remove() 函数
【发布时间】：2015-03-06 14:41:31
【问题描述】：

我已经编写了下面的程序，总的来说它似乎工作正常...除了 remove() 函数不断删除队列中的所有值甚至没有被调用...

#include <iostream>
#include <string>

using namespace std;

//Create a node struct
struct Node {
  int data;
  Node *next;
  Node *prev;
};

class Queue {
private:
  Node *head;
  Node *tail;
  int size;
public:
  Queue();
  ~Queue();
  void add(int d);
  int remove();
  bool isEmpty();
  void printQueue(bool o);
};

//set to NULL
Queue::Queue() {

  head = tail = NULL;
  size = 0;

}

//destructor
//call remove until empty
Queue::~Queue() {

  while (!isEmpty())
    remove();
}

//adds a node with the given data at the back of the queue
void Queue::add(int d) {

  Node *temp = new Node;
  temp->data = d;
  temp->next = NULL;
  temp->prev = tail;

  if (isEmpty()) {

    //add to head
    head = temp;
    tail = temp;

    cout << "Added: " << tail->data << endl;

  } else {

    //append
    tail->next = temp;
    tail = temp;

    cout << "Added: " << tail->data << endl;
  }
  size++;
}

//removes the node at the head of the queue and returns its data
int Queue::remove() {
//TODO DOESNT WORK PROPERLY

  if (isEmpty()) {

    tail = NULL;

    cout << "The queue is empty." << endl;

    return 0;

  } else {

    Node *temp = head;
    int value = head->data;

    cout << "Removed: " << head->data << endl;

    //moves pointer to next node
    head = head->next;

    if (head)
      head->prev = NULL;

    size--;
    delete temp;
    return value;
  }
}

//determines if the queue is empty
bool Queue::isEmpty() {
  return (size == 0);
}

//prints the contents of the queue from front to back, or front
//to back, depending on the value of the parameter
void Queue::printQueue(bool o) {

  if (isEmpty()) {

    cout << "The queue is empty." << endl;

  } else {

    Node *p;

    if (o == true) {

      p = head;

      cout << "Printing front to back:" << endl;

      //print front to back
      while(p != NULL) {
        cout << p->data << " ";
        p = p->next;
      }

      cout << endl;

    } else if (o == false) {

      p = tail;

      cout << "Printing back to front:" << endl;

      //print back to front
      while (p != NULL) {
        cout << p->data << " ";
        p = p->prev;
      }

      cout << endl;
    }
  }
}

int main() {

  Queue q;

  q.add(9);
  q.add(10);
  q.add(11);
  q.add(12);
  q.add(13);
  q.add(14);
  q.add(15);
  q.add(16);

  q.printQueue(true);
  q.printQueue(false);

  return 0;
}

在主程序中，remove() 甚至没有被使用，但是输出是

Added: 9
Added: 10
Added: 11
Added: 12
Added: 13
Added: 14
Added: 15
Added: 16
Printing front to back:
9 10 11 12 13 14 15 16 
Printing back to front:
16 15 14 13 12 11 10 9 
Removed: 9
Removed: 10
Removed: 11
Removed: 12
Removed: 13
Removed: 14
Removed: 15
Removed: 16

为什么这种情况不断发生？我能做些什么来修复它？请帮忙:(

提前致谢！

【问题讨论】：

标签： c++ list function queue nodes

【解决方案1】：

问题实际上可能不在代码中，乍一看它表现正常。

您将Queue q; 声明为main 中的局部变量。这意味着当 main 完成它的执行时， q 超出范围，并调用析构函数。

希望这对此事有所启发......

【讨论】：