Employee Class in C++

Question

For the following code, does anyone understand how and why the friend function is working? what is emp and why does emp.boss point to name?? The purpose was the create a class that adds employees to subteams.• Since employees can have only one boss, if we hire someone who already has a boss, we should report a failure.

#include<iostream>.
#include<string>
#include<vector>

class Employee{

friend ostream& operator<<(ostream& os, const Employee& emp){
    cout<< "Name: "<< emp.name << endl;
    cout<< "Boss: ";
    if (emp.boss != nullptr){
        cout<< emp.boss -> name << endl;
    }
    else
    {
        cout << "I am the boss." <<endl;
    }
    cout<< "Team ..." <<endl;
    for(fize_t i = 0; i < emp.myTeam.size(); i++) {
        cout << "\t" << emp.myTeam[i] -> name << endl;
    }
    return os;
}

private:
    string name;
    Employee* boss;
    vector<Employee*> myTeam;

public:
    Employee(const string& name, Employee* boss = nullptr): name(name),    boss(boss)
//
    {
    if(boss != nullptr){
        boss -> myTeam.push_back(this);
    }
}
//condition check





bool addToTeam(Employee& newGuy)
{
    if(newGuy.boss != nullptr)
    {
        return False;
    }
    newGuy.boss = this;
    myTeam.push_back(&newGuy);
    return true;
}
bool removeFromTeam(Employee& emp)
{
    for(size_t i=0; i <myTeam.size(); i++)
    {
        if(myTeam[i] == &emp) //the first item in the vector is the reference to &emp, so now the first item is now the address of the employee?        {
            emp.boss = nullptr;
            myTeam[i] = myTeam[myTeam.size() -1];
            myTeam.pop_back();
            return true;
        }
    }
    return false;
    }
};

I tried to get the correct code.

`emp.boss` doesn't "point to" `name`. The `->` operator dereferences the `emp.boss` pointer and accesses its `name` member. It is syntactic sugar for `(*emp.boss).name`. — 0x5453, Mar 21 '22 at 18:39
It looks like you need [a good C++ book](https://stackoverflow.com/questions/388242/the-definitive-c-book-guide-and-list). The concepts of passing arguements to functions, references and operator overloading should be in any beginner level book. — Yksisarvinen, Mar 21 '22 at 18:39
So if it accesses the name member, what does it look for?Is nullptr pointing to nothing, but then points to the name to see if the name is the boss?? I am very confused — swindy, Mar 21 '22 at 18:44
I don't really understand what is your issue. Perhaps you should run your debugger, examine values of variables when the function is executed and it will clear up to you? — Yksisarvinen, Mar 21 '22 at 18:51
This question may need more focus. From your comment, you may be misunderstanding what `!=` means. `if (emp.boss != nullptr)` is an if statement that is only entered if `emp.boss` is **not** `nullptr`. _"pointing to nothing, but then points to the name"_ does not describe what this code is saying. — Drew Dormann, Mar 21 '22 at 20:03

score 0 · Answer 1 · edited Mar 22 '22 at 00:59

Okay, let's look at the method:

friend ostream& operator<<(ostream& os, const Employee& emp){
    cout<< "Name: "<< emp.name << endl;
    cout<< "Boss: ";
    if (emp.boss != nullptr){
        cout<< emp.boss -> name << endl;
    }
    else
    {
        cout << "I am the boss." <<endl;
    }
    cout<< "Team ..." <<endl;
    for(fize_t i = 0; i < emp.myTeam.size(); i++) {
        cout << "\t" << emp.myTeam[i] -> name << endl;
    }
    return os;
}

First off, this is an operator<< method. You would use it like this:

Employee someone;
... populate someone with data, however you're doing that

std::cout << someone << std::endl;

Basically, this just makes it easy to print an Employee to any stream -- like cout or a file.

Most of it should make sense. I think you were wondering about this:

    cout<< emp.boss -> name << endl;

This is writing out this employee's boss's name. So if you did this:

Employee emp;
Employee boss;
boss.name = "Fred";

Then that section of code would write out Fred. You can read it left to right:

Start with emp
Then go to the boss field
Then because boss points somewhere (the code checks), look at the boss's name field.

except that `operator<<` is hard coded to output to cout - which is totally wrong — pm100, Mar 22 '22 at 01:00

Employee Class in C++

1 Answers1