将cin的输出分配给变量 [英] Assigning the output of cin to variables

问题描述

我正在尝试学习如何使用 cin 和 getline 编写一个纸质分级程序，我可以在学校使用。对于初学者来说这是一个棘手的项目，但它让我知道了我需要学习的知识，这是我需要做的第一件事。

I'm trying to learn how to use cin and getline to write a paper grading program that I can use at school. It's kind of a tricky project for a beginner but it lets me know what I need to learn and this is the first thing I need to do.

int main()
{
    string grader;
    int x;
    cout << "Who will I be assisting today? ";
    getline (cin, grader);
    cout << "Hello " << grader << ".\n";
    cout << "How manny questions are on the test you will be grading? ";
    getline (cin, x);
    cout << "this is a " << x << "question test graded by" << grader << ".\n";
}

让我说我回答了第一个问题的John Doe，然后回答了第二个问题的20 。我希望它打印这是John Doe分级的20个问题的测试。

我哪里出错了？
我敢肯定这是一个愚蠢的错误，但却困扰着我。我是新手，为您的无知感到抱歉。我将对此程序有更多与用户输入无关的问题。是否可以在此处发布这些问题或开始新话题？
谢谢

Lets say I answered John Doe for the first question, then 20 for question two. I want it to print "this is a 20 question test graded by John Doe"
Where am I going wrong? I'm sure it's a stupid mistake but it's bugging me. I'm a novice so sorry for the ignorance. I will have more questions regarding this program that will not have to do with user in-put. is it ok to post these questions here, or start new topics? thanks

推荐答案

由于您未说明错误原因，而且还可能是缺少include /命名空间。
完整的可运行/可编译程序将是：

Since you did not state what was your error, and alas it could also be a missing include/namespace. The complete runnable/compilable program would be:

#include <iostream>
#include <string>

using namespace std;
int main()
{
    string grader;
    int x;
    cout << "Who will I be assisting today? ";
    getline (cin, grader);
    cout << "Hello " << grader << ".\n";
    cout << "How manny questions are on the test you will be grading? ";
    cin >> x;
    cout << "this is a " << x << "question test graded by" << grader << ".\n";
}

无论如何，这将在您输入问题数量后立即关闭（或从shell / cmd调用您的exe文件）-所以不要怀疑您是否看不到结果。

anyway, this will immediatly close after you entered the amount of questions (or you call your exe from a shell/cmd) - so dont wonder if you cant see the result.

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