c++ - 从 void 子函数检索 int 值到 C++ 中的函数

Question

我是一名开始学习 C++ 编码的基础学生。我正在为我的大学作业做一个调查程序，在测试之后，我发现子函数的总和值不能正确地求和到主函数中的值。任何人的帮助!!!

代码如下:

#include <iostream>

using namespace std;

int Q1();
int Q2();
int Q3();
int Q4();
int Q5();

int main()
{
    char select;
    int E, A, C, N, O;
    int extroversion=0, agreeableness=0, conscientiousness=0, neuroticism=0,          opennesstoexperience=0;

    cout << "This is a Self-Esteem Questionnaire." << endl;
    cout << "\nInsert S to start the test (Q to Quit): ";
    cin >> select;
    select=toupper(select);

    if (select=='S')
    {

    cout << "------------------INSTRUCTIONS-----------------" << endl;
    cout << "For each statement 1-50 mark how much you agree" << endl;
    cout << "with on the scale 1-5, where                   " << endl;
    cout << "1=disagree, 2=slightly disagree, 3=neutral,    " << endl;
    cout << "4=slightly agree and 5=agree                   " << endl;
    cout << "-----------------------------------------------" << endl;

    cout << Q1() << endl;
    extroversion+=E;

    cout << Q2() << endl;
    agreeableness+=A;

    cout << Q3() << endl;
    conscientiousness+=C;

    cout << Q4() << endl;
    neuroticism+=N;

    cout << Q5() << endl;
    opennesstoexperience+=O;

    cout << extroversion << endl;
    cout << agreeableness << endl;
    cout << conscientiousness << endl;
    cout << neuroticism << endl;
    cout << opennesstoexperience << endl;

    }

    else
        if(select=='Q')
   {
        cout << "Program quit!" << endl;
    }
    return 0;
}

int Q1()
{
    int E=0;

    cout << "I am the life of the party." << endl;
    cout << "1=disagree, 2=slightly disagree, 3=neutral," << endl;
    cout << "4=slightly agree and 5=agree" << endl;
    cout << "\nRating: ";
    cin >> E;

    return E;

}

int Q2()
{
    int A=0;

    cout << "I feel little concern for others." << endl;
    cout << "1=disagree, 2=slightly disagree, 3=neutral," << endl;
    cout << "4=slightly agree and 5=agree" << endl;
    cout << "\nRating: ";
    cin >> A;

    return A;

}

int Q3()
{
    int C=0;

    cout << "I am always prepared." << endl;
    cout << "1=disagree, 2=slightly disagree, 3=neutral," << endl;
    cout << "4=slightly agree and 5=agree" << endl;
    cout << "\nRating: ";
    cin >> C;

    return C;

}

int Q4()
{
    int N=0;

    cout << "I get stressed out easily." << endl;
    cout << "1=disagree, 2=slightly disagree, 3=neutral," << endl;
    cout << "4=slightly agree and 5=agree" << endl;
    cout << "\nRating: ";
    cin >> N;

    return N;

}

int Q5()
{
    int O=0;

    cout << "I have a rich vocabulary." << endl;
    cout << "1=disagree, 2=slightly disagree, 3=neutral," << endl;
    cout << "4=slightly agree and 5=agree" << endl;
    cout << "\nRating: ";
    cin >> O;

    return O;

 }`

Answer 1

让我们从简化这个问题开始。

int main()
{
    int E;
    int extroversion=0;

    cout << Q1() << endl;
    extroversion+=E;

    cout << extroversion << endl;
    return 0;
}

int Q1()
{
    int E=0;
    cout << "Give me a number" << endl;
    cin >> E;
    return E;
}

问题是您在 Q1 中声明的变量 E 与您在 主要。因此当你写:

    extroversion+=E;

您正在使用未初始化的变量。解决方案是将 main 重写为:

int main()
{
    int extroversion=0;

    int E = Q1();     // Capture the result of Q1 in E
    cout << E << endl;// ... *then* print it.
    extroversion+=E;  // And now you can use the value of E in this function.

    cout << extroversion << endl;
    return 0;
}

请注意:“减少”问题是您首先应该发布的 - 我们不需要看到大量的文本，我们当然不需要看到您重复做同样的事情五次.删除冗长的文字(并检查问题是否仍然存在)，然后发布减少的问题。