c++ - 如何编辑二进制文件中的特定值？

Question

我有一个航类结构，正在写入二进制文件，我想编辑航类目的地，但我不知道该怎么做。这是我获取用户输入和写入二进制文件的代码。

struct Flight_Details {
    char destination[99];
    char departure[99];
    char time_depart[80];
    char time_arrive[80];
    int flight_number;
};

switch (menu)
        {
            case MENU::NEW_FLIGHT: {
                Flight_Details flight_d;
                cout << "Enter Departure: ";
                cin >> flight_d.destination;
                cout << "Enter Destination: ";
                cin >> flight_d.departure;
                cout << "Enter Departure Time: ";
                cin >> flight_d.time_depart;
                cout << "Please enter arriving time: ";
                cin >> flight_d.time_arrive;
                cout << "Flight Number: ";
                cin >> flight_d.flight_number;

                ofstream file;
                file.open("Flgiht_Details.data", ios::binary);

                if (!file) cout << "Could create/open file";
                else {
                    file.write((char*)&flight_d, sizeof(flight_d));
                    file.close();
                }
                break;
            }
            case MENU::OUTPUT_FILE: {

                ifstream readFile;
                readFile.open("Flgiht_Details.data");
                if (!readFile) cout << "Couldn't open file";
                else {
                    readFile.seekg(0, ios::end);
                    int fileSize = readFile.tellg();
                    int countOfFlights = fileSize / sizeof(Flight_Details);
                    readFile.seekg(0, ios::beg);

                    Flight_Details* flight = new Flight_Details[countOfFlights];
                    readFile.read((char*)flight, countOfFlights *sizeof(Flight_Details));
                    readFile.close();

                    for (int i = 0; i < countOfFlights; i++)
                    {
                        cout << flight[i].destination << "\n" << flight[i].departure << "\n" << flight[i].time_depart << "\n" << flight[i].time_arrive << "\n" << flight[i].flight_number << "\n\n";
                    }
                    break;
                }
            }

            case MENU::EDIT: {
                Flight_Details* flight_d;
                ifstream readFile;
                readFile.open("Flgiht_Details.data");
                if (!readFile) cout << "Couldn't open file";
                else {
                    readFile.seekg(0, ios::end);
                    int fileSize = readFile.tellg();
                    int countOfFlights = fileSize / sizeof(Flight_Details);
                    readFile.seekg(0, ios::beg);

                    Flight_Details* flight = new Flight_Details[countOfFlights];
                    readFile.read((char*)flight, countOfFlights * sizeof(Flight_Details));
                    readFile.close();

                    ofstream file;
                    char edit[50];
                    cout << "Edit: ";
                    cin.getline(edit, 50);

                    for (int i = 0; i < countOfFlights; i++)
                    {
                        if (strcmp(flight_d[i].destination, edit) == 0)
                        {
                            //edit file
                        }
                    }

                }
            }       
        }

Answer 1

由于您的 Flight_Details 在写入时具有固定的字节大小，因此很容易计算每个航类在文件中的位置。一般等式是 flight_number * sizeof(Flight_Details)//其中 flight_number 是 0..N。这将计算要查找的文件中的字节偏移量。从那里开始，只需从文件中准确读取 sizeof(Flight_Details) 字节并将该缓冲区转换为 Flight_Details，在这种情况下，这可能意味着您只需填充每个在 Flight_Details 中手动使用字节缓冲，或者如果您愿意，您可以将 reinterpret_cast() 数据的 char* 转换为正确的类型(只有当您的缓冲区恰好是 sizeof(Flight_Details) 长时，这才是一个不错的主意。通过某种构造函数手动复制字节会更安全。