Question

我对c ++很新，这里有很多代码，所以我会尽力将它压缩到问题区域。当我尝试使用getline获取用户输入时获取此错误。由于我不希望文件名中的空格（我制作文件），我使用cin＆lt;＆lt;哪个工作正常，但在尝试读取文件时遇到了同样的错误。代码如下

// includes here

using namespace std;

//other prototypes here
string getUserDataFromFile(vector<int>&, int&, string);


int main()
{
    vector<int> numbers;
    numbers.reserve(50);
    int numberOfElements = 0;
    int number = 0;
    int numToFind = 0;
    int numberPosition = -1;
    int useFile = 0;
    string filename = "";
    string fileReadMessage = "";
    string output = "";
    string outFilename = "";


    cout << "Would you like to load the data from a file?(1 for yes 0 for no)";
    cin >> useFile;
    cin.ignore(INT_MAX, '\n');
    //get user data for manual input
    if(useFile == 0)
    {
        //code here for manual input(works fine)...
    }
    //get userdata for file input
    else
    {
        cout << "Please Enter the file path to be opened" << endl;

        //fixed after adding cin.ignore(INT_MAX, '\n'); 
        //see next function for another problem
        getline(cin, filename);

        fileReadMessage = getUserDataFromFile(numbers, numToFind, filename);
    }

    //some code to get data for output


    return 0;
}




//function to get user data from file
//@param v(vector<int>&) - vector of integers.
//@param numToFind(int&) - the number we are looking for
//@param filename(string) - the filename of the file with data
//@return message(string) - a message containing errors or success.
string getUserDataFromFile(vector<int>& v, int& numToFind, string filename)
{
    string message = "File Accepted";
    string line = "";
    int numOfElements = 0;
    int count = 0;

    ifstream fileToRead(filename.c_str());

    //using 'cin >>' in main, the program runs till here then breaks

    //if message is a file, extract message from file
    if (fileToRead.is_open())
    {
        while (getline(fileToRead,line))
        {
            //code to do stuff with file contents here
        }
        fileToRead.close();
    }
    else
    {
        message = "Unable to open file.";
    }
    return message;
}

我在故障区域留下了一些评论，遗漏了大部分我没有遇到过麻烦或无法测试的代码。任何帮助表示赞赏。谢谢！

所以我的第一个问题是通过添加cin.ignore（INT_MAX，'\ n'）修复的;对下一个问题的任何猜测？它是下一个函数中的行if（fileToRead.is_open（））

Answer 1

添加

cin.ignore();

之前：

getline(cin, filename);

否则，输入useFile后输入的输入将被读入filename。

std :: out_of_range在getline的内存位置错误

1 个答案: