Question

我有一个名为 TeamLeader ， ProductionWorker 和员工

的课程

ProductionWorker 扩展了班级员工。

TeamLeader 扩展 ProductionWorker 。有问题的构造函数如下：

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

错误内容如下：没有匹配的构造函数用于初始化＆＃39; ProductionWorker＆＃39;

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

我在课程 ProductionWorker 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;
}

如果我添加＆＃34;缺少＆＃34;有问题的 TeamLeader 构造函数的参数如此

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

}

我收到以下错误： TeamLeader.cpp：23：128：错误：＆＃39; employeeName＆＃39;是＆＃39; Employee＆＃39;

的私人会员

对于 TeamLeader

无法访问的其他两个参数，也会发生此错误

谁能告诉我如何解决这个问题？因为如果感觉像一个永无止境的圈子......

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拥有姓名等，那么TeamLeader构造函数必须接受名称：

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列表，而不是类体内的赋值语句。

如果您希望团队负责人没有姓名（虽然我不知道您将如何在此方法中设置名称）并且: ProductionWorker(shift, hourlyPayRate)工作，那么您需要添加一个构造函数给{ {1}}有两个参数，例如：

ProductionWorker

NB。此答案假定ProductionWorker :: ProductionWorker(int shift, double hourlyPayRate) : shift(shift), hourlyPayRate(hourlyPayRate) { }和shift是hourlyPayRate的成员变量。

如果您使用的是C ++ 11，那么您可以使用委派构造函数来避免重复这么多。另外，查看默认参数。

我不能在不违反c ++扩展规则的情况下制作匹配的构造函数

1 个答案: