C++ C++；计算工资问题

您好，我正在尝试制作一个程序，在给定员工的工资和工作绩效时，该程序将计算员工奖金，尽管每次运行代码时，无论我输入的工资是多少，奖金都计算为1。我必须保留我创建的函数，它是我家庭作业的一部分。如果您能提供有关该问题的任何指导，我们将不胜感激

#include <iostream>
using namespace std;
double bonus(double salary,int a);
int main()
{
    double salary;
    enum jobp{poor = 0,average = 1,good = 2};
    jobp performance;
    int a = performance;
    cout<<"Enter your salary and job performance (as a 0 for poor,1 for average and 2 for good)s$
    cin>>salary>>a;
    bonus(salary,a);
    cout<<"your bonus is "<<bonus;
    return 0;
}
double bonus(double salary, int a)
{
    double bonus;
if (a == 2)
{
    double c;
    c = .10;
    bonus = salary * c;
    return bonus;
}
else if (a  == 1)
{
    double b;
    b = .05;
    bonus = salary * b;
    return bonus;
}
else
{
    return 0;
}
}

#包括
使用名称空间std；
双倍奖金（双倍工资，INTA）；
int main（）
{
双薪；
枚举jobp{poor=0，average=1，good=2}；
作业绩效；
int a=性能；
cout嗨，这应该能帮你解决问题。
更改此项：

cout<<"Enter your salary and job performance (as a 0 for poor,1 for average and 2 for good)s$
cin>>salary>>a;
bonus(salary,a);
cout<<"your bonus is "<<bonus;

cout捕获返回奖金的价值

#include <iostream>
using namespace std;
double bonus(double salary,int a); //this is for function right?
int main()
{
    double salary;
    double bonus; //create a variable for bonus
    enum jobp{poor = 0,average = 1,good = 2};
    jobp performance;
    int a = performance;
    cout<<"Enter your salary and job performance (as a 0 for poor,1 for average and 2 for good)s$
    cin>>salary>>a;
    bonus = bonus(salary,a); //catch the return
    cout<<"your bonus is "<<bonus;
    return 0;
}

#包括
使用名称空间std；
双倍奖金（双倍工资，整数）；//这是功能的权利？
int main（）
{
双薪；
双倍奖金；//为奖金创建一个变量
枚举jobp{poor=0，average=1，good=2}；
作业绩效；
int a=性能；
无法尝试让用户分别输入工资和绩效

cout << "Please enter your salary: " << endl;
cin >> salary;
cout "Please  job performance (as a 0 for poor,1 for average and 2 for good)" << endl;
cin >> a;

cout工资；
cout“请注意工作表现（0分表示较差，1分表示一般，2分表示良好）”>a；
您的奖金函数返回一个值，您可能需要从那里开始。
cout << "Please enter your salary: " << endl;
cin >> salary;
cout "Please  job performance (as a 0 for poor,1 for average and 2 for good)" << endl;
cin >> a;