程序跳过第二个cin

时间:2013-10-15 20:06:38

标签: c++ cout cin

我正在制作一个C ++ Mind Reader程序,该程序已接近完成。但是,感觉需要跳过第二个cin。我搜索过,我不确定是什么问题。我已经检查了代码,我打赌我做了一些愚蠢的事,但我仍然对此感到困惑。跳过的cin在第32行,这是我的代码:

#include <iostream>
#include <string>
#include <cstdlib>

using namespace std;

int main()
{
    //declaring variables to be used later
    string name;
    string country;
    int age;

    //header goes below
    cout << "                     @@@@@@@@@@@@-MIND READER-@@@@@@@@@@@@\n\n";

    //asks if the user would like to continue and in not, terminates
    cout << "Would like you to have your mind read? Enter y for yes and n for no." << endl;
    cout << "If you do not choose to proceed, this program will terminate." << endl;
    string exitOrNot;
    //receives user's input
    cin >> exitOrNot;
    //deals with input if it is 'y'
    if (exitOrNot == "y"){
        cout << "Okay, first you will need to sync your mind with this program. You will have to answer the following questions to synchronise.\n\n";

        //asks questions
        cout << "Firstly, please enter your full name, with correct capitalisation:\n\n";
        cin >> name;

        cout << "Now please enter the country you are in at the moment:\n\n";
        cin >> country; //<------ Line 32

        cout << "This will be the final question; please provide your age:\n\n";
        cin >> age;

        //asks the user to start the sync
        cout << "There is enough information to start synchronisation. Enter p to start the sync...\n\n";
        string proceed;
        cin >> proceed;
        //checks to see if to proceed and does so
        if (proceed == "p"){
            //provides results of mind read
            cout << "Sync complete." << endl;
            cout << "Your mind has been synced and read.\n\n";
            cout << "However, due to too much interference, only limited data was aquired from your mind." << endl;
            cout << "Here is what was read from your mind:\n\n";

            //puts variables in sentence
            cout << "Your name is " << name << " and you are " << age << " years old. You are based in " << country << "." << endl << "\n\n";

            cout << "Thanks for using Mind Reader, have a nice day. Enter e to exit." << endl;
            //terminates the program the program
            string terminate;
            cin >> terminate;
            if (terminate == "e"){
                exit(0);
            }

        }

    }
    //terminates the program if the input is 'n'
    if (exitOrNot == "n"){
        exit(0);
    }

    return 0;
}

编辑:这是我运行时会发生的事情:

                     @@@@@@@@@@@@-MIND READER-@@@@@@@@@@@@

Would like you to have your mind read? Enter y for yes and n for no.
If you do not choose to proceed, this program will terminate.
y
Okay, first you will need to sync your mind with this program. You will have to
answer the following questions to synchronise.

Firstly, please enter your full name, with correct capitalisation:

John Smith
Now please enter the country you are in at the moment:

This will be the final question; please provide your age:

13
There is enough information to start synchronisation. Enter p to start the sync.
..

p
Sync complete.
Your mind has been synced and read.

However, due to too much interference, only limited data was aquired from your m
ind.
Here is what was read from your mind:

Your name is John and you are 13 years old. You are based in Smith.


Thanks for using Mind Reader, have a nice day. Enter e to exit.
e

Process returned 0 (0x0)   execution time : 78.220 s
Press any key to continue.

这是一个更清晰的屏幕截图:http://puu.sh/4QZb3.png 我不能在这篇文章中附上它,因为我没有足够的代表。

还值得注意的是它如何将用户的姓氏用作国家/地区。 我认为这个问题与输入不是整数有关。

感谢。

4 个答案:

答案 0 :(得分:5)

您只能在cin中输入一个单词。相反,使用getline(cin, string name);如果仍然无效,请在cin.ignore();之前添加getline(cin, string name);,如下所示:

string country;
cout << "Now please enter the country you are in at the moment:\n\n";
cin.ignore();
getline(cin, country);

现在肯定会有用。

答案 1 :(得分:3)

问题在于您使用:cin >>从用户处获取字符串。如果用户输入的单词超过1个,则会导致代码中的行相互跳过。要解决此问题,我们将使用:getLine(cin,yourStringHere)从用户处获取字符串。这是你的代码所有修复:

#include <iostream>
#include <string>
#include <cstdlib>

using namespace std;

int main()
{
    string name;
    string country;
    int age;
    cout << "                     @@@@@@@@@@@@-MIND READER-@@@@@@@@@@@@\n\n";

    cout << "Would like you to have your mind read? Enter y for yes and n for no." << endl;
    cout << "If you do not choose to proceed, this program will terminate." << endl;
    string exitOrNot;
    getline (cin,exitOrNot); /*<-----Changed from cin to getLine*/
    if (exitOrNot == "y"){
        cout << "Okay, first you will need to sync your mind with this program. You will have to answer the following questions to synchronise.\n\n";

        cout << "Firstly, please enter your full name, with correct capitalisation:\n\n";
        getline(cin,name); /*<--Another string*/

        cout << "Now please enter the country you are in at the moment:\n\n";
        getline(cin,country); /*<--Another string*/

        cout << "This will be the final question; please provide your age:\n\n";
        cin >> age;

        cout << "There is enough information to start synchronisation. Enter p to start the sync...\n\n";
        string proceed;
        cin >> proceed;
        if (proceed == "p"){
            cout << "Sync complete." << endl;
            cout << "Your mind has been synced and read.\n\n";
            cout << "However, due to too much interference, only limited data was aquired from your mind." << endl;
            cout << "Here is what was read from your mind:\n\n";

            cout << "Your name is " << name << " and you are " << age << " years old. You are based in " << country << "." << endl << "\n\n";

            cout << "Thanks for using Mind Reader, have a nice day. Enter e to exit." << endl;
            string terminate;
            cin >> terminate;
            if (terminate == "e"){
                exit(0);
            }

        }

    }
    if (exitOrNot == "n"){
        exit(0);
    }

    return 0;
}

答案 2 :(得分:0)

上述用户所说的,cin只允许将1个字输入每个cin的字符串。因此,我认为你想要的是:getline(cin,name)

答案 3 :(得分:-1)

cin.get();
cout << "Now please enter the country you are in at the moment:\n\n";
cin >> country; //<------ Line 32

在cout之前键入cin.get(),如上所述,您的问题将得到解决。