C 将数据读取到一个结构，但它跳过了我的一个获取线，使该元素空白

c++ Reading data to a struct but it is skipping one of my getlines making the element a blank

本文关键字：一个 获取 我的 空白 元素 过了 数据 读取 结构      更新时间：2023-10-16

这是我的代码，我并没有真正看到这里的问题，我敢肯定它很简单，我只需要另一对新鲜的眼睛才能找到。

#include <iostream>
#include <string>
using namespace std;
//Struct
struct MovieData
{
    string title;
    string director;
    int year;
    int runTime;
    double productionCost;
    double fYRevenue;
};
//Prototypes
MovieData getMovieData();
void printMovieData(MovieData );
int main()
{
    //Variables
    MovieData m1, m2;
    cout << "Enter data for movie 1:";
    m1 = getMovieData(); //call get data
    cout << "Enter data for movie 2:";
    m2 = getMovieData(); // Call get data
    printMovieData(m1); //Call print
    printMovieData(m2); //Call print
    return 0;
}
/*
Passes a struct of type movie data and 
collects data needed to print out 
later on
*/
MovieData getMovieData()
{
    MovieData m;
    char line;
    cout << "nnWhat is the title of the movie?: ";
    getline(cin, m.title);
    cout << "nWho was the director of the movie?: ";
    getline(cin, m.director);
    cout << "nWhat year was the movie made?: ";
    cin >> m.year;
    cout << "nHow long is the movie in minutes?: ";
    cin >> m.runTime;
    cout << "nHow much did it cost to produce this movie?: ";
    cin >> m.productionCost;
    cout << "nHow much did the movie make in its first year?: ";
    cin >> m.fYRevenue;
    cout << "n";
    return m;
}
/*
Passes a variable of type movie data
then prints out the information inside
*/
void printMovieData(MovieData m)
{
    cout << "nnThe movie data for " << m.title << " is as follows.n"; 
    cout << "Title: " << m.title;
    cout << "nDirector: " << m.director;
    cout << "nYear Made: " << m.year;
    cout << "nRunning Time: " << m.runTime;
    cout << "nProduction cost: " << m.productionCost;
    cout << "nRevenue: " << m.fYRevenue ;
}

，如果你们能找到问题，我非常感谢它弄乱了一会儿，我在这里无法理解这个问题。第二次GetMoviedata被称为第一个标题是空白行。但是，当我把它们当作cin＆lt;＆lt;M.Title;当我输入两个部分时，它跳到了整个功能，与董事名称Carl Simons一样，洪水也会跳过整个过程。

不要忘记清除您的缓冲区！

它不是完美的，但它是您与...

一起工作的开始 
MovieData getMovieData()
{
    MovieData m;
    char line;
    //you have to clear out your buffer here
    cin.ignore();
    cin.sync();
    cout << "nnWhat is the title of the movie?: ";
    getline(cin, m.title);
    cout << "nWho was the director of the movie?: ";
    getline(cin, m.director);
    cout << "nWhat year was the movie made?: ";
    cin >> m.year;
    cout << "nHow long is the movie in minutes?: ";
    cin >> m.runTime;
    cout << "nHow much did it cost to produce this movie?: ";
    cin >> m.productionCost;
    cout << "nHow much did the movie make in its first year?: ";
    cin >> m.fYRevenue;
    cout << "n";
    return m;
}

您在getMovieData函数中混合了cin >>和getline的使用。这导致了令人困惑的行为，因为cin >>不会从输入缓冲区消耗尾随的新线（当您阅读第一部电影的收入时），而getline并不希望看到缓冲区中的尾随newline（当您阅读标题时下一部电影）。

将getline用于所有用户输入，此问题将消失。如果您需要读取一个数字，请与getline一起使用CC_7或其他数字转换功能。例如：

std::string input;
getline(cin, input);
m.fYRevenue = stoi(input);

如果您阅读了几个整数，则可以将上述行包裹到其自己的函数中。

尝试：

MovieData getMovieData()
{
    MovieData m;
    char line;
    cout << "nnWhat is the title of the movie?: ";
    cin>> m.title;
    cout << "nWho was the director of the movie?: ";
    cin >> m.director;
    cout << "nWhat year was the movie made?: ";
    cin >> m.year;
    cout << "nHow long is the movie in minutes?: ";
    cin >> m.runTime;
    cout << "nHow much did it cost to produce this movie?: ";
    cin >> m.productionCost;
    cout << "nHow much did the movie make in its first year?: ";
    cin >> m.fYRevenue;
    cout << "n";
    return m;
}
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