c++中LinkedList调用时的显示函数

Question

我试图理解链接列表，但我试图复制的每个例子都给了我一个分割错误。这是我正在编写的示例程序。

在contacts.h

struct people{
  string last;
  string first;
  int age;
  people *next;
};
class Contacts {
 private:
  people *head;
 public:
  Contacts() {
   head=NULL;
  };
  void add(string, string, int);
  void display();
};

在main.cpp

//menu items
//cin >> option
//if option == 1, addContact
//if option == 8, displayContact
//basic in and out here no need to show code

在main.h

void addContact() {
  Contacts *ptr;     
  int i, loop=0, a=0;
  string l, f;
  cout << "number of contacts " << endl;
  cin >> loop;
  for(i=0; i<loop; i++) {
    cout << "enter last name ";
    cin >> l;
    cout << "enter first name ";
    cin >> f;
    cout << "enter age ";
    cin >> a;
  }
  ptr->add(l,f,a);
}
void display() {
 Contacts *ptr;
 ptr->display();
}    

在contacts.cpp

void Contacts::add(string l, string f, int a) {
  people *node = new people;
  node->last=l;
  node->first=f;
  node->age=a;
  node->next=NULL;
}
void Contacts::display() {
 people *tmp = head;
 while(tmp!=NULL) {
   cout << tmp->last << endl;
   cout << tmp->first << endl;
   cout << tmp->age << endl;
   tmp->next;
 }

的添加函数工作然后display()给出一个段错误

Answer 1

add成员函数应按如下方式定义

void Contacts::add( const string &l, const string &f, int a ) 
{
    head = new people { l, f, a, head };
}

或者如果编译器不支持带有new操作符的初始化列表，则

void Contacts::add( const string &l, const string &f, int a ) 
{
    people *p = new people;
    p->last = l;
    p->first = f;
    p->age = a;
    p->next = head;
    head = p;
}

成员函数display应该声明为

void Contacts::display() const;

，定义为

void Contacts::display() const
{
    for ( people *tmp = head; tmp; tmp = tmp->next )
    {
        cout << tmp->last << endl;
        cout << tmp->first << endl;
        cout << tmp->age << endl;
        cout << endl;
    }
}

其他两个函数应按以下方式定义

void addContact( Contacts &contacts ) 
{
    int n = 0;
    cout << "number of contacts " << endl;
    cin >> n;
    for ( int i = 0; i < n; i++ ) 
    {
        string l, f;
        int a = 0;
        cout << "enter last name ";
        cin >> l;
        cout << "enter first name ";
        cin >> f;
        cout << "enter age ";
        cin >> a;
        contacts.add( l, f, a );
    }
}

void display( const Contacts &contacts ) 
{
    contacts.display();
}    

在main中定义一个类型为

的对象

Contacts contacts;

并将其用作上述函数的参数