Question

我正在尝试格式化链表，以便在每行上打印5个节点。我不知道如何做到这一点，因为我是操作员重载的新手。这是我尝试过的东西，但我陷入了困境，似乎无法掌握这个概念

     ostream &operator<<(ostream &os, List &s){ 

     nodeType<Type>* current = s.head; 
     int i = 0; 

 while (current != NULL) //while more data to print 
 { 
 os << current->info << " "; 
 current = current->link; 

 ++i; 

 if (i % 5 == 0) { 
     cout << '\n'; 
     i = 0; 
 } 
 } 

os << '\n'; // print the ending newline 

       return os; 
     }

其余代码

list.cpp

List::List()
{
node *head = NULL;
node *precurrent = NULL;
node *current = NULL;
int *temp = 0;
insert = 0;
search = 0;
remove = 0;
}

List::~List()
{
while (head != 0)
    remove();
}

void List::insert(int insert)
{
if (head==null)    \\If there is no list already, create a new head.
{
    temp = new Node;
    temp->data = insert;
    head = temp;
}
else                \\otherwise, insert the new node after current
{
    temp = new Node;
    temp->data = insert;
    temp->next = current->next;
    current->next = temp;
}
}

void List::search(int search)
{
current=head;
while (head->next != 0)  //Cycle through the list, and if the number is found, say so
{
    if (current->data = search)
        cout<<"Number found."<<endl;
    else
        cout<<"Number not found."<<endl;
}
}

void List::remove(int remove)
{
if (head == null)
    cout <<"Error.  No List."<<endl;
else if (head->next == null)
{
    num = head->data;
    delete head;
    head=null;
    current=null;
}
else if (head == current)
{
    temp = head->next;
    num = head->data;
    delete head;
    head=temp;
    current=temp;
}
else
{
    temp = current->next;
    num = current->data;
    delete current;
    precurrent->next = temp;
    current = temp;
}
}

list.h

//CLASS PROVIDED:  list                                 
//
// CONSTRUCTOR for the list class:
//   list()
//     Description:     Constructor will initialize variables
//     Preconditions:   None
//     Postcondition:   int insert = ""
//                      int search = ""
//                      int remove = ""
//   ~list()
//     Description:     Destructor destroys variables
//     Preconditions:   None
//     Postcondition:   variable deleted
//
// MEMBER FUNCTIONS for the list class      
//
//   string insert(int)
//     Description: Inserts an integer into a linked list   
//     Precondition: none
//     Postcondition: function returns Success/Error message.
//
//   string search(int);
//     Description:     Searches for certain linked list member and returns int to set current variable
//     Precondition:    none
//     Postcondition:   function returns int
//
//    string remove(int);
//     Description:     removes linked list member
//     Precondition:    user sends int to be deleted
//     Postcondition:   function returned string sddress
//   
//   void display(void);
//     Description:     displays entire linked list
//     Precondition:    none
//     Postcondition:   function returns screen output
//
//    void quit(void);
//     Description:     closes program
//     Precondition:    none
//     Postcondition:   none
//


#ifndef EMPLOYEE_H
#define EMPLOYEE_H

#include <string>
#include <iostream>
#include <cstdlib>

using namespace std;

class list
{     
  public:
//CONSTRUCTOR/DESTRUCTOR---------------------
    list();
    ~list();                                     

//GETS---------------------------------------
    void insert(int);
    string search(int);
    string remove(int);
    void display(void);
    void quit(void);

  private:

        int insert;
        int search;
        int remove;  


};




#endif

Answer 1

您需要将current设置为s.head，而不仅仅是head，这是未定义的，因为此非成员运算符重载（顾名思义）不是成员。

你也在推动指针完全错误;你应该在每次迭代中打印一个info，如下所示：

编辑：如果你想每行打印5个，那么这样做：

int i = 0;

while (current != NULL) //while more data to print
{
     os << current->info << " ";
     current = current->link;

     if (i % 5 == 0) {
         cout << '\n';
         i = 0;
     } else
         ++i;
}

os << '\n'; // print the ending newline

此外，Type未定义（除非它在您未发布的代码中的某处）。如果您的List是模板，则还需要让运算符重载模板。

请初始化变量而不是声明它们然后分配给它们。这样：

nodeType<Type> *current; //pointer to traverse the list
current = head; //set current so that it points to the first node

应该是

nodeType<Type>* current = s.head;

通过重载格式化链接列表输出＆lt;＆lt;

1 个答案: