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如何在getline之后输入

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#include <iostream>
#include <string>
using namespace std;

struct Student
{
    int ID;
    long phno;
    string name;
    string depart;
    string email;
};

int main ()
{
    Student S1 ;

    cout << "\n=======================================================\n" ;

    cout << "Enter ID no. of student 1       : " ; cin >> S1.ID ;
    cout << "Enter name of student 1         : " ; getline(cin, S1.name) ; cin.ignore();
    cout << "Enter department of student 1   : " ; getline(cin, S1.depart) ; cin.ignore();
    cout << "Enter email adress of student 1 : " ; getline(cin, S1.email) ; cin.ignore();
    cout << "Enter phone number of student 1 : " ; cin >> S1.phno ;     

    return 0;
}

问题是它没有在电子邮件地址后输入，程序忽略了在phno中输入，直接在emailadress后退出 .

c++ getline

1 回答

  • 0

    我对你的代码做了一些修改 .

    请注意，我直接在变量（ cin >> S1.ID 或 cin >> S1.phno ）上使用 cin 后才调用 cin.ignore() .

    这是因为当你在int上使用 cin 时，它会将 \n 留在缓冲区中 . 当你以后打电话给 getline(cin,...) 时，你只是把剩下的 \n 吸了，这就算是你的整个"line."

    有一个工作示例here .

    #include <iostream>
#include <string>
using namespace std;

struct Student
{
    int ID;
    long phno;
    string name;
    string depart;
    string email;
};

int main ()
{
    Student S1 ;

    cout << "\n=======================================================\n" ;

    cout << "Enter ID no. of student 1       :\n" ;
    cin >> S1.ID ; 
    cin.ignore();

    cout << "Enter name of student 1         :\n" ;
    getline(cin, S1.name) ; 

    cout << "Enter department of student 1   :\n" ;
    getline(cin, S1.depart) ;

    cout << "Enter phone number of student 1 :\n" ;
    cin >> S1.phno ;
    cin.ignore();

    cout << "Enter email adress of student 1 :\n" ;
    getline(cin, S1.email) ;    

    cout << endl << endl;

    cout << S1.ID << endl << S1.name << endl << S1.depart << endl << S1.phno << endl << S1.email << endl;

    return 0;
}
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