我无法在不违反c++扩展规则的情况下创建匹配的构造函数

Question

我有一个名为ProductionWorker TeamLeader，Employee的类

ProductionWorker扩展类Employee。

扩展ProductionWorker.的TeamLeader所述的构造函数如下：

TeamLeader :: TeamLeader(int trainingHoursCompleted, int shift, double hourlyPayRate) : ProductionWorker(shift, hourlyPayRate){
    monthlyBonus = 1000;
    requiredTrainingHours = 20;
    this->trainingHoursCompleted = trainingHoursCompleted;
}

错误读取如下：没有用于初始化“ProductionWorker”的匹配构造函数

...shift，double hourlyPayRate)：ProductionWorker(shift，hourlyPayRate)。

ProductionWorker类中的构造函数如下：

ProductionWorker :: ProductionWorker() : Employee(){
    shift = 0;
    hourlyPayRate = 0;
}

ProductionWorker :: ProductionWorker(int shift, double hourlyPayRate, string employeeName, string hireDate, int employeeNumber) : Employee(employeeName, hireDate, employeeNumber) {
    this->shift = shift;
    this->hourlyPayRate = hourlyPayRate;
}

如果我将“缺失”参数添加到TeamLeader构造函数中，如下所示

    TeamLeader :: TeamLeader(int trainingHoursCompleted, int shift, double hourlyPayRate) : ProductionWorker(shift, hourlyPayRate, employeeName, hireDate, employeeNumber){
monthlyBonus = 1000;
requiredTrainingHours = 20;
this->trainingHoursCompleted = trainingHoursCompleted;

}

I得到以下错误： TeamLeader.cpp:23:128: error：'employeeName‘是'Employee’的私有成员

对于TeamLeader无法访问的其他两个参数，也会发生此错误。

有人能告诉我怎么解决这个问题吗？因为如果感觉像是一个永无止境的圆圈..。

TeamLeader.cpp

#include <stdio.h>
#include <String>
#include "TeamLeader.h"

using namespace std;



TeamLeader :: TeamLeader() : ProductionWorker(){
    monthlyBonus = 1000;
    requiredTrainingHours = 20;
    trainingHoursCompleted = 0;
}

TeamLeader :: TeamLeader(int trainingHoursCompleted, int shift, double hourlyPayRate) : ProductionWorker(shift, hourlyPayRate){
    monthlyBonus = 1000;
    requiredTrainingHours = 20;
    this->trainingHoursCompleted = trainingHoursCompleted;
}

void TeamLeader :: setTrainingHoursCompleted(int trainingHoursCompleted){
    this->trainingHoursCompleted = trainingHoursCompleted;
}

ProductionWorker.cpp

#include "ProductionWorker.h"

ProductionWorker :: ProductionWorker() : Employee(){
    shift = 0;
    hourlyPayRate = 0;
}

ProductionWorker :: ProductionWorker(int shift, double hourlyPayRate, string employeeName, string hireDate, int employeeNumber) : Employee(employeeName, hireDate, employeeNumber) {
    this->shift = shift;
    this->hourlyPayRate = hourlyPayRate;
}


void ProductionWorker :: setShift(int shift){
    this->shift = shift;
}

void ProductionWorker :: setHourlyPayRate(double hourlyPayRate){
    this->hourlyPayRate = hourlyPayRate;
}

Employee.cpp

Employee :: Employee(){
    employeeName = "NO NAME ENTERED";
    hireDate = "NO DATE ENTERED";
    employeeNumber = 0;
}


Employee :: Employee(string employeeName, string hireDate, int employeeNumber){
    this->employeeName = employeeName;
    this->hireDate = hireDate;
    this->employeeNumber = employeeNumber;
}

void Employee :: setEmployeeName(string employeeName){
    this->employeeName = employeeName;
}

void Employee :: setHireDate(string hireDate){
    this->hireDate = hireDate;
}

void Employee :: setEmployeeNumber(int employeeNumber){
    this->employeeNumber = employeeNumber;
}

Answer 1

如果您希望团队负责人有一个名称等等，那么TeamLeader构造函数必须接受这个名称：

TeamLeader :: TeamLeader(int trainingHoursCompleted, int shift, double hourlyPayRate, string employeeName, string hireDate, int employeeNumber) 
    : ProductionWorker(shift, hourlyPayRate, employeeName, hireDate, employeeNumber)
    , monthlyBonus(1000), requiredTrainingHours(20)
    , trainingHoursCompleted(trainingHoursCompleted)
{ }

注意:最好使用构造函数初始化程序列表，而不是类主体中的赋值语句。

如果您希望团队负责人没有名称(虽然我不知道您将如何在这种方法中设置名称)并让: ProductionWorker(shift, hourlyPayRate)工作，那么您需要向ProductionWorker添加一个构造函数，该构造函数包含两个参数，例如：

ProductionWorker :: ProductionWorker(int shift, double hourlyPayRate) : 
    shift(shift), hourlyPayRate(hourlyPayRate)
{ }

注意：这个答案假设shift和hourlyPayRate是ProductionWorker的成员变量。

如果您使用的是C++11，那么您可以使用委托构造函数来避免重复多次。此外，查看默认参数。