问仅在第二次访问链接列表中的结构后才进行分割
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Stack Overflow用户

提问于 2017-11-27 15:53:36

回答 1查看 38关注 0票数 1

抱歉，标题不清楚，我真的不知道该怎么描述这个问题。我正在学计算机科学的第一年，所以我对C++还不太了解。然而，试图查找这一问题并没有帮助。

问题:在主要功能中，"printRawData“朋友函数被调用了两次。函数应该打印由类"LinkedList“存储的链接列表中的每个元素。它第一次工作，但第二次我得到一个分割错误。我真的不知道我做错了什么。我的T.A.说他认为结构的字符串变量"element_name“在被访问时被破坏了。

如果我没有很好地解释我的问题，或者如果我破坏了任何一种堆叠式的礼节，那么很抱歉我的代码太乱了。我很感激我得到的任何帮助。

//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);
}