使用C 中的链接列表的文件处理

问题：我试图写（二进制写作）双链接列表的对象到文件＆amp;也从中写下。我必须编写对象的完整内容，然后将其从文件加载，然后将其存储到新对象中以按FIFO顺序重新创建列表。我以为我写作正确，但我真的不知道如何从文件加载（阅读）。

记住：我只是想保存和阅读节点的CONTENTS，＆amp;NOT POINTERS.

代码：

//template type BOOK class
template<class mytype>
class BOOK      
{
private:
    static int count;   //declaration of static variable to set ID for books
public:
    BOOK<mytype> *next, *prev;  //BOOK type pointers; 'next' to store address of 
next BOOK & 'prev' to store address of previous BOOK
    int ID;         //variable to store ID of a book
    string bookName;//string to store name of a book
    string author;  //string to store name of author of book
    string book_type;//string to store type of a book
    long copies;    //variable to store no. of copies a book
    long price;     //variable to store price of a book
    string status;  //to store status of a book, either its in stock or not
    dynamicQueue<string> book_queue;    //created an object of queueClass as data member of each Book
    BOOK()  //Constructor 0 argument to initialize everything
    {
        count++;    //increment counter
        ID=count;   //assign counter to ID to be ID of newly added book
        next = prev = 0;        //Initializing both pointers to 0
        bookName = "";
        author = "";
        book_type = "";
        copies = price = 0;
        status= "InStock";
    }
    BOOK(BOOK *n =  0, BOOK *p = 0, string book = "", string athr = "", string buk_type = "", long cp=0, long pr=0) //Constructor multiple arguments, to store information about a book
    {
        next = n;       //store contents of user-given value n into next
        prev = p;       //store contents of user-given value p into previous
        bookName = book;//store contents of user-given value book into bookName
        author = athr;  //store contents of user-given value athr into author
        book_type = buk_type;//store contents of user-given value buk_type into book_type
        copies = cp;    //store contents of user-given value cp into copies
        price = pr;     //store contents of user-given value pr into price
        status= "InStock";
        count++;        //increment counter
        ID=count;       //assign counter to ID to be ID of newly added book
    }
};
template <class mytype>    // declaration of
int BOOK<mytype>::count=0; // static variable to set ID for books
//--------------------

添加新书的主要部分。

BookStoreDataBase<char> obj;    //created object of Doubly linked list
string Book, Author, Booktype;
long Copies=1, Price=0;
cout<<"Enter Name of Book = ";  cin>>Book;
cout<<"Enter Author = ";        cin>>Author;
cout<<"Enter Type of Book = ";  cin>>Booktype;
cout<<"Enter Number of Copies = ";  cin>>Copies;
cout<<"Enter Price (PKR) = ";   cin>>Price;
obj.addBook(Book, Author, Booktype, Copies, Price);

保存功能以将所有数据保存到文件

template <class mytype>
void DoublyLinkedList<mytype>::save_data()
{
    NODE<mytype> * temp = head; //made copy of head
    fstream file;     //created new file
    file.open("mydata.txt", ios::binary | ios::out | ios::app);
    while(temp->next!=0) //Until link list end
    {
          file.write(reinterpret_cast<char*>(&temp), sizeof(temp));
          temp = temp - > next; //move temp to next node
    }
    file.write(reinterpret_cast<char*>(&temp), sizeof(temp)); //write again for last  
                                                              //book's data
    file.close();
}

现在，我几乎不知道如何从文件中读取列表，将内容存储到每个节点＆amp;在干扰保存的安排的情况下，按FIFO顺序重新创建了列表。这样我稍后可以打印。我已经练习了很多，去了论坛等，但找不到任何具体解决方案。请帮助我。预先感谢

我的努力样本

template <class mytype>
void DoublyLinkedList<mytype>::load_data()
{
    fstream file;
    file.open("mydata.txt", ios::binary | ios::in);
    while(!file.eof())
    {
        NODE<mytype> *temp = new NODE<mytype>;
        file.read(reinterpret_cast<char*>(&temp), sizeof(temp));
        if(is_Empty())
        {
            head = tail = temp;
        }
        else
        {
            temp->prev->next = temp;
            temp->next=0;
        }
    }
    file.close();
}
//-------------------

无编译时间错误。

运行时错误： in Dobulylist.exe中的0x00CD8391未经处理的异常：0xc00000055：访问违规写作位置0x00000004。