使用C ++中的链接列表进行文件处理

时间:2012-11-01 15:30:01

标签: c++ file-io binaryfiles file-handling doubly-linked-list

问题:我正在尝试将(二进制写入)双重链接列表的对象写入文件&也写出来。 我必须编写对象的完整内容,然后从文件中加载它,并将其存储到新对象中,以FIFO顺序重新创建列表。 我认为我写的正确,但我真的不知道如何从文件中加载(读取)它。

请记住:我只是想保存并阅读节点的CONTENTS,& NOT POINTERS.

CODE:

//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 = "\0";
        author = "\0";
        book_type = "\0";
        copies = price = 0;
        status= "InStock";
    }

    BOOK(BOOK *n =  0, BOOK *p = 0, string book = "\0", string athr = "\0", string buk_type = "\0", 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和放大器中。在扰乱保存的安排的情况下,按FIFO顺序重新创建列表。所以我以后可以打印出来。我已经练习了很多,去了论坛等,但没有找到任何具体的解决方案。请帮帮我。提前致谢


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();
}
//-------------------

没有编制时间错误。

RUNTIME ERROR: DobulyList.exe中0x00CD8391处的未处理异常:0xC0000005:访问冲突写入位置0x00000004。

1 个答案:

答案 0 :(得分:0)

我认为你最好的选择是在BOOK类上有一个方法,它可以以预定义的方式序列化内容。

template<class mytype>
class BOOK      
{
private:
    //members
public:
    //more members

    ....

    bool read( istream& in );
    bool write( ostream& out );

    ....

};

因为你想要二进制文件 - 你需要为ID编写和int,然后是字符串的大小,字节等...

然后你反过来阅读。读取一个int并分配给ID,然后读取大小,读取大小字符并分配给字符串等...

e.g

//fill in template stuff I'm lazy...
bool Book::read( istream& in )
{
     size_t stringSize; //you need to think about what types you read/write
     char buffer[SomeArbitrarySize];
     file.read( ID ,sizeof(int));
     file.read( stringSize ,sizeof(size_t));
     file.read( buffer, stringSize );
     mStringMember = buffer;
     ... etc ...
}

然后对列表中的每个项目委派调用以读/写节点。您可能还希望首先为节点数写入/读取条目。

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>;
        temp->read(file);
        if(is_Empty())
        {
            head = tail = temp;
        }
        else
        {
            temp->prev->next = temp;
            temp->next=0;
        }
    }
    file.close();
}