c++ - 仅在第二次访问链表中的结构后出现段错误

Question

抱歉标题不清楚，我真的不知道如何描述这个问题。我是计算机科学的第一年，所以我对 C++ 还不是很了解。但是，尝试查找此问题并没有帮助。

问题:在主函数中，“printRawData”友元函数被调用了两次。该函数应该打印由“LinkedList”类存储的链表的每个元素。它第一次工作，但第二次出现段错误。我真的不知道我做错了什么。我的助教说他认为结构的字符串变量“element_name”在访问时被破坏了。

如果我没有很好地解释我的问题，或者如果我违反了任何一种 stackoverflow 礼节，那么对于困惑的代码，我们深表歉意。我感谢我得到的任何帮助。

//Note: C++ 11 is needed, due to to_string use
#include <iostream>
#include <string>

using namespace std;

struct Node {
  string element_name;
  int element_count;
  Node* next;
};

class LinkedList{
  private:
    Node* first;
  public:
    LinkedList();
    ~LinkedList();
    bool isEmpty();
    void AddData(string name, int count);
    friend void printRawData(LinkedList l);
};

//where the error occurs
void printRawData(LinkedList l){
  Node* n = l.first;
  while (n != NULL) { //iterates through the linked list and prints each element
    cout << n->element_name << " : " << n->element_count << endl;
    n = n->next;
  }
}

LinkedList::LinkedList(){
  first = NULL;
}

LinkedList::~LinkedList(){
  Node* n = first;
  while (n != NULL) {
    Node* temp = n;
    n = temp->next;
    delete temp;
  }
}

bool LinkedList::isEmpty(){
  return first == NULL;
}

void LinkedList::AddData(string name, int count){
  Node* newnode = new Node;
  newnode->element_name = name;
  newnode->element_count = count;
  newnode->next = NULL;

  Node* n = first;

  //if the linked list is empty
  if(n == NULL){
    first = newnode;
    return;
  }

  //if there's only one element in the linked list,
  //if the name of first element comes before the name of new element,
  //first element's pointer is to the new element.
  //otherwise, the new node becomes the first and points to the previous first
  //element.
  if (n->next == NULL){
    if (n->element_name < newnode->element_name){
      n->next = newnode;
      return;
    } else {
      newnode->next = first;
      first = newnode;
      return;
    }
  }

  //if the first element's name comes after the new element's name,
  //have the new element replace the first and point to it.
  if (n->element_name > newnode->element_name){
    newnode->next = first;
    first = newnode;
    return;
  }

  //iterating through linked list until the next element's name comes after
  //the one we're inserting, then inserting before it.
  while (n->next != NULL) {
    if (n->next->element_name > newnode->element_name){
      newnode->next = n->next;
      n->next = newnode;
      return;
    }
    n = n->next;
  }

  //since no element name in the linked list comes after the new element,
  //the node is put at the back of the linked list
  n->next = newnode;
}



main(){
  LinkedList stack;

  stack.AddData("Fish", 12);
  stack.AddData("Dog", 18);
  stack.AddData("Cat", 6);

  printRawData(stack);
  printRawData(stack);
}

Answer 1

函数 void printRawData(LinkedList l) 按值传递参数，因此它获得了 LinkedList 对象的拷贝。

但是，拷贝包含 first 指针的拷贝，但不复制任何节点。所以当这个拷贝被销毁时，LinkedList 析构函数将删除所有节点。

然后原件损坏了。

您可能希望传递引用而不是创建拷贝。

---

这也是 std::list 具有执行“深复制”的复制构造函数和赋值运算符的原因，其中节点也被复制(不仅仅是列表头)。