设置对象属性时C++内存访问冲突

Question

我的代码中有一部分无法工作。构建一个基本的链接列表来学习指针。我想我已经把它的大部分都写下来了，但是任何使用我创建的函数(push_back)的尝试都会在为指针设置值时抛出内存访问错误。

不完全确定出了什么问题，因为使用push_front很好，它的工作原理几乎完全相同。

有什么想法吗？

代码：

driver.cpp

#include <string>
#include <iostream>
#include "linklist.h"
#include "node.h"

using namespace std;

// printList function
// Purpose: Prints each node in a list
// Returns: None.
// Pre-Conditions: List must have member nodes.
// Post-Conditions: None.
void printList(linklist);

int main()
{
     linklist grocery;

     grocery.push_front(new node("milk", "1 gallon"));
     grocery.push_front(new node("bread","2 loaves"));
     grocery.push_front(new node("eggs","1 dozen"));
     grocery.push_front(new node("bacon","1 package"));
     cout << "First iteration:" << endl;
     printList(grocery);
     cout << "----------------------" << endl << endl;

     grocery.push_front(new node("hamburger","2 pounds"));
     grocery.push_front(new node("hamburger buns", "1 dozen"));
     cout << "Second iteration:" << endl;
     printList(grocery);
     cout << "----------------------" << endl << endl;

     node* deleteMe = grocery.pop_front();
     delete deleteMe;
     cout << "Third iteration:" << endl;
     printList(grocery);
     cout << "----------------------" << endl << endl;

     grocery.push_back(new node("orange juice","2 cans"));
     grocery.push_back(new node("swiss cheeese","1 pound"));
     cout << "Fourth iteration:" << endl;
     printList(grocery);
     cout << "----------------------" << endl << endl;

     deleteMe = grocery.pop_back();
     delete deleteMe;
     cout << "Fifth iteration:" << endl;
     printList(grocery);
     cout << "----------------------" << endl << endl;

     while (grocery.getNodeCount() != 0)
     {
          deleteMe = grocery.pop_front();
          cout << "Cleaning: " << deleteMe->getDescription() << endl;
          delete deleteMe;
     }

     system("PAUSE");
     return 0;
}

void printList(linklist input)
{
   node* temp = input.getFirst();
   for (int i = 0; i < (input.getNodeCount()); i++)
   {
      cout << temp->getQuantity() << " " << temp->getDescription() << endl;

      temp = temp->getNextNode();
   }
}

node.h

#pragma once
#include <string>

using namespace std;

class node
{
public:
// Default Constructor
// Values, "none", "none", NULL.
node();

// Parameterized Constructor
// nextNode initialized NULL and must be explicitly set.
node(string descriptionInput, string quantityInput);

// getDescription function
// Purpose: Returns node description.
// Returns: string
// Pre-Conditions: None.
// Post-Conditions: None.
string getDescription();

// setDescription function
// Purpose: Sets node description
// Returns: Void
// Pre-Conditions: None
// Post-Conditions: None
void setDescription(string);

// getQuantity function
// Purpose: Returns node quantity.
// Returns: string
// Pre-Conditions: None.
// Post-Conditions: None.
string getQuantity();

// setQuantity function
// Purpose: Sets node quantity
// Returns: Void
// Pre-Conditions: None
// Post-Conditions: None
void setQuantity(string);

// getNextNode function
// Purpose: Returns pointer to next node in list sequence.
// Returns: node pointer
// Pre-Conditions: None.
// Post-Conditions: None.
// Note: Not set during initialization. Must be explicitly done.
node* getNextNode();

// setNextNode function
// Purpose: Sets pointer to next node in list sequence.
// Returns: None.
// Pre-Conditions: None.
// Post-Conditions: None.
// Note: Not set during initialization. Must be explicitly done.
void setNextNode(node*);
private:
string description;
string quantity;
node* nextNode;
};

node.cpp

#include "node.h"


node::node()
  :description("none"),
  quantity("none"),
  nextNode(NULL)
  {}

node::node(string descriptionInput, string quantityInput)
  :description(descriptionInput),
  quantity(quantityInput),
  nextNode(NULL)
  {}

string node::getDescription()
{
   return description;
}

void node::setDescription(string descriptionInput)
{
   description = descriptionInput;
}

string node::getQuantity()
{
   return quantity;
}

void node::setQuantity(string quantityInput)
{
   quantity = quantityInput;
}

node* node::getNextNode()
{
   return nextNode;
}

void node::setNextNode(node* input)
{
   nextNode = input;
}

linklist.h

#pragma once
#include "node.h"

class linklist
{
public:
// Constructor
// Builds an empty list
linklist();

// push_front function
// Purpose: Takes node pointer. Places that node at beginning of list.
// Returns: None
// Pre-Conditions: None
// Post-Conditions: None
void push_front(node*);

// pop_front function
// Purpose: Removes first node from list.
// Returns: Node pointer. NODE IS NOT DESTROYED.
// Pre-Conditions: List must have a node to remove.
// Post-Conditions: Node is not destroyed.
node* pop_front();

// getFirst function
// Purpose: Returns node pointer to first node in list
// Returns: node pointer
// Pre-Conditions: List must have a node added.
// Post-Conditions: None.
node* getFirst();

// push_back function
// Purpose: Takes node pointer. Places that node at end of list.
// Returns: None
// Pre-Conditions: None
// Post-Conditions: None
void push_back(node*);

// pop_back function
// Purpose: Removes last node from list.
// Returns: Node pointer. NODE IS NOT DESTROYED.
// Pre-Conditions: List must have a node to remove.
// Post-Conditions: Node is not destroyed.
node* pop_back();

// getNodeCount function
// Purpose: Returns nodeCount
// Returns: int
// Pre-Conditions: None.
// Post-Conditions: None.
int getNodeCount();
private:
node* firstNode;
node* lastNode;
int nodeCount;
};

linklist.cpp

#include "linklist.h"


linklist::linklist()
    :firstNode(NULL),
    lastNode(NULL),
    nodeCount(0)
{}

void linklist::push_front(node* input)
{
    node* temp = getFirst();

    input->setNextNode(temp);

    firstNode = input;
    nodeCount++;
}

node* linklist::pop_front()
{
    node* temp = getFirst();

    firstNode = temp->getNextNode();

    nodeCount--;
    return temp;
}

node* linklist::getFirst()
{
    return firstNode;
}

void linklist::push_back(node* input)
{
    node* temp = lastNode;

    temp->setNextNode(input);

    lastNode = temp;
    nodeCount++;
}

node* linklist::pop_back()
{
    node* oldLast = lastNode;
    node* temp = firstNode;

    // find second to last node, remove it's pointer
    for (int i = 0; i < (nodeCount - 1); i++)
    {
        temp = temp->getNextNode();
    }
    temp->setNextNode(NULL);

    lastNode = temp;

    nodeCount--;
    return oldLast;
}

int linklist::getNodeCount()
{
    return nodeCount;
}

Answer 1

根据node的定义，这是一个单链列表。(否则，你也必须包含prevNode)。

因此--操作列表的“回”是不平凡的。要弹出列表的“回”，您需要横过整个列表并识别新的“最后”元素。

你确定你做得对吗？包括处理所有的“极端”案例(比如删除最后一个元素等等)？

最好能发布push_back和pop_back的代码。

也许你没有在你的push_front和pop_front中正确地设置push_front和pop_front。除非你试图操纵你的“背部”，否则你可能不会注意到这一点。

同时发布push_front和pop_front的代码。

编辑

好吧，我看到密码了。还有很多错误。

void linklist::push_front(node* input)
{
    node* temp = getFirst();

    input->setNextNode(temp);

    firstNode = input;
    nodeCount++;

    // Missing:
    if (!temp)
        lastNode = firstNode;
}


node* linklist::pop_front()
{
    node* temp = getFirst();

    firstNode = temp->getNextNode();

    // Missing:
    if (!firstNode)
        lastNode = 0;

    nodeCount--;
    return temp;
}


void linklist::push_back(node* input)
{
    node* temp = lastNode;

    // instead of
    // temp->setNextNode(input);
    // lastNode = temp;

    // should be:
    if (temp)
        temp->setNextNode(input);
    else
        firstNode = input;

    lastNode = input;

    nodeCount++;
}

node* linklist::pop_back()
{
    node* oldLast = lastNode;

    if (firstNode == lastNode)
    {
        firstNode = 0;
        lastNode = 0;
    } else
    {
        node* temp = firstNode;

        // find second to last node, remove it's pointer
        for (int i = 0; i < (nodeCount - 1); i++)
        {
            temp = temp->getNextNode();
        }
        temp->setNextNode(NULL);

        lastNode = temp;
    }

    nodeCount--;
    return oldLast;
}

Answer 2

推方法有错误，因为当你在前面推的时候，你不能控制这个元素是否也是最后一个，当你在最后推的时候也是相似的。它导致你没有连接整个列表，因为开始部分不知道结束部分。我希望这是可以理解的。

另外，pop方法也是错误的--同样的问题。您无法控制列表是否为空

Answer 3

当列表为空时，在push_back temp中为NULL。因此，temp->setNextNode(input)失败。您需要区分空列表的特殊情况。

顺便问一下，如果您允许在后面和前面的操作，也许您需要一个双链接的列表？否则，您需要遍历整个列表(可能很大)，才能弹出最后一个元素，因为您没有指向前一个元素的“直接”链接。

顺便说一句，你的操作是德克的，不是名单的。