Question

我正在尝试从payroll.dat输入此数据：

40.0     10.00
38.5      9.50
16.0      7.50
42.5      8.25
22.5      9.50
40.0      8.00
38.0      8.00
40.0      9.00
44.0     11.75

进入类payroll个对象的2d数组。

以下是我错误尝试的内容：

int count = 0;
const int numEmploy = 7;            // Number of employees
const int col = 2;                  // Number of categories
Payroll empData[numEmploy][col];    // Array to hold objects
ifstream inputFile;

    inputFile.open("payroll.dat");
    if (!inputFile)
        cout << "There was an error opening the file. Please make sure it exists." << endl;
    else
    {
        while(count < numEmploy && inputFile >> empData[][])
        {
            inputFile >> empData[count][0] >> empData[count][1];
            count++;
        }
    }

最终，我需要将payroll.dat中的值设为[0][0]位置，然后乘以[0][1]，然后乘以[1][0]*[1][1]，再乘以[2][0]*[2][1]等。将结果显示为总薪酬。

我认为我对>>运算符的理解是关闭的。在这种情况下，这是流提取，按位或技术上两者？

我对正在发生的事情的叙述的理解是：只要count小于numEmploy并且empData[][]从payroll.dat接收值，请插入第一个可用的值将payroll.dat中的数据块放入empData[0][0]，将第二个块（第一个块右侧的块）插入empData[0][1]。然后循环返回并将以下块（下一行，payroll.dat的第一列）插入empData[1][0]，然后将右边的块插入empData[1][1]。继续此操作，直到count大于或等于numEmploy。然后每个empData[#][#]将成为类Payroll的对象。至少这是我想要的：P

甚至可以这样做吗？还是我坚持使用两个不同的阵列？

到目前为止，这是我的整个代码：

class Payroll
{
    private:

        double payRate;      // holds an employee hourly pay rate

        double hoursWorked;  // an employee's hours worked

    public:

        Payroll()  // empty constructor sets the payRate and hoursWorked to zero
        {
            payRate = hoursWorked = 0;
        }

        Payroll(double payR, double hoursW) //constructor checks for payR and hoursW to be positive
                    // and sets payRate and hours worked; sets to zero if negative values are provided
        {
            if (payR < 0) payR = 0;
            if (hoursW < 0) hoursW = 0;
        }

        void setPayRate(double payR) //mutator for payRate; checks for payR to be positive or sets to zero
        {
            payRate = payR;
        } 

        void setHoursWorked(double hoursW) //mutator for hoursWorked; checks for positive hoursW or sets to zero
        {
            hoursWorked = hoursW;
        }

        double getPayRate() //accessor to return payRate
        {
            return payRate;
        }

        double getHoursWorked() // accessor to return hoursWorked
        {
            return hoursWorked;
        }

        double getGrossPay() // computes and returns gross pay including OVERTIME, if any
        {
            float normHours, overHours, grossPay;

            if (hoursWorked > 40)
            {
                overHours = (hoursWorked - 40);
                normHours = (hoursWorked - overHours);
                grossPay = (overHours * payRate * 1.5) + (normHours * payRate);
            }

            return grossPay;

        }
};





int main ()
{
int count = 0;
const int numEmploy = 7;            // Number of employees
const int col = 2;                  // Number of categories
Payroll empData[numEmploy][col];    // Array to hold objects
ifstream inputFile;

inputFile.open("payroll.dat");
if (!inputFile)
    cout << "There was an error opening the file. Please make sure it exists." << endl;
else
{
    while(count < numEmploy && inputFile >> empData[][])
    {
        inputFile >> empData[count][0] >> empData[count][1];
        count++;
    }
}

inputFile.close();

cout << fixed << showpoint << setprecision(2) << endl;

cout << "Employees' gross pay:" << endl;

for (int index = 0; index < numEmploy; index++)
{

    empData[index][2].setPayRate();
    empData[index][1].setHoursWorked();
    cout << "Employee # " << index + 1 << empData.getGrossPay();

}


    return 0;
}

Answer 1

我想你想要更像这样的东西，

Payroll empData[numEmploy] ;

...

while ( count < numEmploy )
{
  inputFile >> empdata[count].payRate >> empdata[count].hoursWorked ;
  count ++ ;
}

然后你的工资只是empdata[x].payRate * empdata[x].hoursworked

如果读取循环不在Payroll的成员函数中，那么您必须使用setter：

while ( count < numEmploy )
{
  double tmppay, tmphours ;
  inputFile >> tmppay >> tmphours ;
  empdata[count].setPayRate( tmppay ) ;
  empdata[count].setHoursWorked( tmphours) ;
  count ++ ;
}

从具有两列的文件加载2D数组

1 个答案: