C++ 没有构造函数的实例与参数列表匹配。

Question

我无法弄清楚我的代码有什么问题。我正在尝试完成 edx 的实验室作业，但我无法弄清楚为什么我的代码无法运行。该作业需要创建一个 Student 类、Teacher 类和 Course 类。我正在尝试找出 Course 类构造函数，但它不会接受我的 Teacher 对象。我使用一个指针来创建对象，以便为它分配内存空间(我在 edx 上学到的东西之一，如果我误解了请告诉我)。于是尝试在构造函数中使用reference，还是不行。这是代码。我使用“未知”作为填充词。

主要代码:

#include "stdafx.h"
#include <iostream>
#include <string>
#include "Student.h"
#include "Teacher.h"
#include "Course.h"

using namespace std;

int main()
{
    Teacher *teach = new Teacher("Jane", "DoeDoe", 25, "Unknown", "Unknown", "Unknown");
    Student *stud1 = new Student();
    Student *stud2 = new Student("John", "Doe", 19, "Unknown", "Unknown", "Unknown");
    Student *stud3 = new Student("Jane", "Doe", 23, "Unknown", "Unknown", "Unknown");
    //Method had two postions. First and Last
    stud1->setName("Unknown", "Unknown");
    stud1->setAge(20);
    stud1->setAddress("Unknown");
    stud1->setCity("Unknown");
    stud1->setPhone("Unknown");

    Course *c = new Course("Intermediate C++", teach, stud1, stud2, stud3);


    return 0;
}

类(class).h

#pragma once

#include <iostream>
#include "Student.h"
#include "Teacher.h"
#include <string>

using namespace std;

class Course {
private:
    string name;
    Teacher teacher;
    Student student1;
    Student student2;
    Student student3;


public:


    Course(string n, Teacher &t, Student &s1, Student &s2, Student &s3);

    ~Course();


};

类(class).cpp

#include "stdafx.h"
#include "Course.h"



Course::Course(string n, Teacher &t, Student &s1, Student &s2, Student &s3)
{
    name = n;
    teacher = t;
    student1 = s1;
    student2 = s2;
    student3 = s3;

}

Course::~Course()
{
}

Answer 1

Course(string n, Teacher &t, Student &s1, Student &s2, Student &s3);

希望您将引用传递给 t、s1、s2..

Course *c = new Course("Intermediate C++", teach, stud1, stud2, stud3);

这里你传递了teach，这是一个teacher*。你想要

Course *c = new Course("Intermediate C++", *teach, *stud1, *stud2, *stud3);

您可能想阅读有关 references and pointers 的更多信息