Hello, can anyone helpe with the following:
rr.getNext().doWorkDuringTimeSlice(); I´ve copied this code from professional C++ second edition which raised: how can a method from class Process to be called from a object from class RoundRobin? page 411.
I´ll be gratefull for the answer. Tanks a lot, Robson

Hi,

rr is of type RoundRobin<Process>.
Calling getNext() on rr returns a reference to a Process on which you can then call doWorkDuringTimeSlice().

Marc

MARC, Thank you so much, Robson

Hello. This is about page 42 and 46. Your first useful program.

I don´t believe the method addEmployee needs to return a reference.
I set this function to return void and I also commented out the return code
in that method. I tested all cases in the user interface at runtime though the
program worked properly.

Am I right?

code
void Database::addEmployee(string inFirstName, string inLastName)
{
Employee theEmployee;
theEmployee.setFirstName(inFirstName);
theEmployee.setLastName(inLastName);
theEmployee.setEmployeeNumber(mNextEmployeeNumber+ +);
theEmployee.hire();
mEmployees.push_back(theEmployee);

// return mEmployees[mEmployees.size()-1];
}
/code

page 46

code
void doHire(Database& inDB)
{
string firstName;
string lastName;

cout << "First name? ";
cin >> firstName;
cout << "Last name? ";
cin >> lastName;

try {
inDB.addEmployee(firstName, lastName);
} catch (const std::exception&) {
cerr << "Unable to add new employee!" << endl;
}
}
/code

It is not required that a method like add return a reference to the object that it has added but if it does you can continue to do operation on the reference after adding it


I.e.

DB.AddEmployee("bill", "bob").Age = 23;

Her I have added an employee and set his age all in one line of code

mmorgan30 is correct.
It's not required to return a reference, but client code could use it to immediately call another method on the new employee.