覆盖操作符<<在c ++ [英] Overriding of operator<< in c++

问题描述

我正在使用C ++为我的学校制作一个专案

I'm working on a project for my school in C++

我有两个类别：Employe和Teacher。
从Employe派生的老师已覆盖其职能。

I have 2 class : Employe and Teacher. Teacher derived from Employe and has overrides of his functions.

我们覆盖运算符<< 打印雇员或教师的一些信息。
每个类都有一个 const int属性LevelAcces _ 。
Employe 为 5 ，教师为 20 。

We override the operator << to print some information of the Employes or Teachers. Each class has a const int attribute LevelAcces_. For Employe, it's 5 and for Teacher it's 20.

当我在main.cpp中创建一个Teacher时，我调用了operator<<的老师打印他的信息。
所以这个函数被调用：

when I create an Teacher in my main.cpp, I call the override of operator<< of Teacher to print his information. So this function is called :

ostream& operator<<(ostream& os, const Teacher& pTeacher){
    os << (pTeacher);
    return os;
}

但是，函数自身 os<<（pTeacher）;和会导致堆栈溢出的循环。

我想让行os

I want that the line "os << (pTeacher)" calls the operator<< of my class Employe and not of my class Teacher.

覆盖操作符<< in Employe：

Override of operator<< in Employe :

ostream& operator<<(ostream& os, const Employe& pEmploye){
    os << "Name: " << pEmploye.name_ << endl;
    os << "Class: " << pEmploye.getClass() << endl;
    os << "LevelAcces: " << pEmploye.getLevelAccess() << endl;
    return os;
}

我试图将我的老师转为Employe，但是当它打印消息时，LevelAcces是5（我想要20，因为我的Employe是一个老师）。

I tried to cast my Teacher into Employe but when it prints the message, LevelAcces is 5 (and I want 20, because my Employe is a Teacher).

我也试图使用Employe :: operator<但是操作员<<不是Employe的成员，所以它不工作...

I also tried to use Employe::operator<< but operator<< is not a member of Employe so it doesn't work...

所以，这里是我的问题：

So, here is my question :

如何使用我的运算符<< of Employe in my operator<<

我也在想虚拟，但我们的教授告诉我们

I was also thinking of "virtual" but our professor tells us that there is not need to use this word.

提前感谢：）

这是一个更完整的代码：

Here is a more complete code :

main.cpp：

Teacher Garry("Garry");
cout << Garry << endl;

Employe.cpp：

Employe.cpp :

#include "Employe.h"

using namespace std;

Employe::Employe(){
    name_ = "";
}

Employe::Employe(string pName){
    name_ = pName;
}

string Employe::getName() const{
    return name_;
}

unsigned int Employe::getLevelAccess() const{
    return levelAccess_;
}

string Employe::getClass() const{
    return typeid(*this).name();
}

ostream& operator<<(ostream& os, const Employe& pEmploye){
    os << "Name: " << pEmploye.name_ << endl;
    os << "Class: " << pEmploye.getClass() << endl;
    os << "LevelAcces: " << pEmploye.getLevelAccess() << endl;
    return os;
}

在Employe.h中：

With this in Employe.h :

private:
    static const unsigned int LevelAccess_ = 5;

Teacher.cpp：

Teacher.cpp :

#include "teacher.h"
using namespace std;

Teacher::Teacher(string pName){
    nom_ = pName;
}

unsigned int Teacher::getLevelAccess() const{
    return(Employe::getLevelAccess() + accessTeacher_); 
}
string Teacher::getClass() const{
    return typeid(*this).name();
}

ostream& operator<<(ostream& os, const Teacher& pTeacher){
        os << (pTeacher);
        return os;
}

这是Teacher.h：

With this is Teacher.h :

static const unsigned int accesTeacher_ = 15;

推荐答案

我要做的是：define only一个

What I'd do is the following: define only one

ostream& operator<<(ostream& os, const Employe& pEmploye)
{
    return pEmploye.display(os);
}

作为层次结构的基础，
在其中调用受每个派生类覆盖的受保护成员函数 virtual display（），并且显示被委派给它。这有时称为NVI（非虚拟接口）成语。它的工作原理如下：

for the base of the hierarchy, in which you call a protected member function virtual display() that is overridden by each derived class and to which the display is being delegated. This is sometime call the NVI (non-virtual interface) idiom. It works like this:

class Employee
{
    // ...
    friend ostream& operator<<(ostream& os, const Employee& pEmployee)
    {
        return pEmployee.display(os);
    }
protected:
    virtual ostream& display(ostream& os) const
    {
        // implement it here
        return os;
    }
};

class Teacher: public Employee
{
    // ....
protected:
    ostream& display(ostream& os) const override
    {
        // implement it here
        return os;
    }
};

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