c++ - 程序没有提供所需的输出

Question

我有一项任务，我应该执行以下操作。

1. 编写一个程序，询问员工的薪水和服务年限。
2. 如果员工的服务年限超过 10 年。他们得到 10% 的加薪。在 5 到 10 年之间，加薪幅度为 5%。其他人将获得 2% 的加薪。
3. 员工的年度奖金。每 2 年服务 500 美元。
4. 以清晰专业的方式显示计算的信息。

我写了下面的代码作为分配问题的解决方案；但是我总是得到错误的输出。我试图调试程序找出问题所在，我看了书，看了看网络，但我找不到问题所在。

   #include <iostream>
    #include <iomanip>


using std::cout;
using std::cin;
using std::setw;
using std::setprecision;
using std::fixed;
using std::right;
using std::left;
using std::setfill;

//function prototypes
void GetInput(double & salary, int years_service);
void CalcRaise(double & salary, int years_service);
int CalcBonus(int years_service);
void PrintCalculations(int years_service, double salary, int bonus);

int main() 
{
    //variable diclerations
    double salary = 0.00;
    int years_service = 0;
    int bonus = 0;

    //function calls
    GetInput(salary,years_service);
    CalcRaise(salary,years_service);
    CalcBonus(years_service);
    PrintCalculations(years_service,salary,bonus);

    cout << "\n";
    return 0;
}

//prompts the user for input
void GetInput(double  &salary, int years_service)
{
    cout << "Enter Salary: ";
    cin >> salary;
    cout << "\nEnter years of service: ";
    cin >> years_service;
}

//calculates the raise
void CalcRaise(double & salary, int years_service)
{
    double raise = 0.00;


    if (years_service > 10)
    {
        raise = salary * (10 / 100);
        salary = salary + raise;
    }
    else if (years_service >= 5 && years_service <= 10)
    {
        raise = salary * (5 / 100);
        salary = salary + raise;
    }
    else
    {
        raise = salary * (2 / 100);
        salary = salary + raise;
    }
}

//calculates the bonus
int CalcBonus(int years_service)
{
    int bonus = 0;

    bonus = (years_service / 2) * 500;

    return bonus;
}

//outputs the results of the calculations
void PrintCalculations(int years_service, double salary, int bonus)
{
    cout << setw(18) << left << "Years of Service" << setw(16) << left << " Salary after raise" 
        << setw(8) << left << " Bonus\n";
    cout << setw(15) << setfill(' ') << right<< years_service 
        << setw(18) << setfill(' ') << right << salary << setw(8) << right << setfill(' ') << bonus;
}

=========================

The Output:

Enter Salary: 4000

Enter years of service: 7
Years of Service   Salary after raise Bonus
               0              4000       0

Answer 1

错误在这里:

void GetInput(double  &salary, int years_service)

您将 years_service 作为值传递。这将简单地复制您之前拥有的 0。然后你永远不会读它，而是写给它。原来的值当然不会被覆盖。

您应该改为通过引用传递。

void GetInput(double  &salary, int& years_service)