c++ - 当我使用 getline( ) 时变量没有正确读入

Question

我正在尝试为我拥有的 C++ 教科书编写一个简单的地址簿练习程序。出于某种原因，当我尝试使用 getline(cin, variablename, '\n') 将文本存储到字符串时，它不会存储任何内容。

我在 Linux mint 17.1 上使用 g++ 编译器进行编译非常感谢任何帮助。

到目前为止，这是我的代码:

#include <iostream>
#include <string>

using namespace std;

struct peopleData
{
    string name;
    string address;
    string phoneNumber;
};

using namespace std;


peopleData getData (string name, string address, string phoneNumber)
{
    peopleData person;
    person.name = name;
    person.address = address;
    person.phoneNumber = phoneNumber;
    return person;
}


int main ()
{
    int amount_of_people;
    string name;
    string address;
    string phoneNumber;

    cout << "How many people are you entering information for? ";
    cin >> amount_of_people;

    peopleData people[amount_of_people];

    for ( int i = 0; i < amount_of_people; i++)
    {
        cout << "What is person " << i + 1 << "'s name?\n";
        getline( cin, name, '\n' );
        cin.ignore();
        cout << "What is " << name << "'s address?\n";
        getline( cin, address, '\n' );
        cin.ignore();
        cout << "What is ";
        cout << name << "'s phone number?\n";
        getline( cin, phoneNumber, '\n' );
        cin.ignore();
        people[i] = getData(name, address, phoneNumber); 
    }

    cout << "Your address book is finished.\n";
    for ( int x = 0; x < amount_of_people; x++)
    {
        cout << people[x].name 
             << "'s address is "  << people[x].address 
             << "\nand their phone number is " << people[x].phoneNumber << endl;
    }

}

Answer 1

好的，更新一下:你应该在 getline 之前调用 cin.ignore() 而不是在 getline 之后。如果你想按照你现在的方式去做，你在阅读数字后忘记调用 cin.ignore() ，导致第一个看起来完全是空的。

前面的回答是错误的，但下面的说法仍然成立:

你的 '\n' 是多余的。参见 http://www.cplusplus.com/reference/string/string/getline/以供引用。