Question

＆＃34;您的任务是提示用户输入磁盘上路径的文件名。如果文件在指定位置不存在，则程序应退出并显示相应的错误消息＆＃34; 大家好，所以这就是我遇到问题的地方，我能够让用户输入这样的文件名..

1.         cout<< "enter the data file name you wish to open";
2.         cin >>file;
3.         indata.open(file.c_str());
4.         outdata.open(file.c_str());

问题的第二部分是如果文件不存在，程序应该出错，我该怎么做呢，说我的文件名是txt.source，但是用户输入lil.pol怎么办我说它有一个错误，或者换句话说我如何使文件名成为计算机唯一接受的文件名？

Answer 1

您可以尝试打开该文件，如果无法打开，请通过std::cerr发出消息，如下所示：

std::ifstream indata(file.c_str());

if(!indata) // failed to open
{
    std::cerr << "Error: could not open file: " << file << std::endl;
    return 1; // error code
}

// use indata here (its open)

Answer 2

编辑：

读/写数据：

void createfile()
{
    ofstream file_handle("test.txt");
    if (!file_handle)
        return;

    //add record:
    file_handle << "firstname1" << endl;
    file_handle << "lastname1" << endl;
    file_handle << "college1" << endl;
    file_handle << "1001" << endl;

    //add another record:
    file_handle << "firstname2" << endl;
    file_handle << "lastname2" << endl;
    file_handle << "college2" << endl;
    file_handle << "1002" << endl;

    //remember each record is 4 lines, each field is single line
    //this is the file format
}

int main()
{
    createfile();

    ifstream fin("test.txt");
    if (!fin)
    {
        cout << "file not found" << endl;
        return 0;
    }

    ofstream fout("out.txt");//note, it's a different name than input file
    if (!fout)
    {
        cout << "cannot create new file" << endl;
        return 0;
    }

    char buffer[1000];

    while (fin)
    {
        cout << "attempting to read record:\n";
        for (int i = 0; i < 4; i++)
        {
            fin.getline(buffer, 1000, '\n');
            if (!fin) break;

            cout << buffer << endl;//write to screen
            fout << buffer << endl;//write to file

            if (i == 3)
            {
                //buffer is expected to be a number!
                int number = atoi(buffer);
                //multiply by random number 2, just testing
                cout << number * 2 << endl; 
            }
        }
    }

    return 0;
}

如果文件名错误，只需创建一个循环并请求新条目。

int main()
{
    ifstream indata;
    string fname;
    for (;;)
    {
        cout << "enter fname, zero to exit\n";
        cin >> fname;
        if (fname == "0")
            return 0;
        indata.open(fname);
        if (indata)
            break;//file is valid and has been opened now
        cout << "file not found, try again\n";
    }

    return 0;
}

让用户输入文件C ++

2 个答案: