I have a base class where the overriding of an << operator is made.
friend std::ostream& operator<<(std::ostream& out, const Animal& animal);
ostream & operator<<(ostream & out, const Animal& animal)
{
out << animal.age << "," << animal.size << endl;
}
This is in the .h of my Abonne class. I have a inherited class of this Abonne class named Etudiant. I wish to override the << operator in the Etudiant class again, but I want it to call the operator<< of the base class Abonne in the definition of the Etudiant class, plus add more. I tried this:
friend ostream& operator<<(ostream& out, const Dog& dog);
ostream& operator<<(ostream& out, const Dog& dog)
{
return out << dog << endl;
}
It's not working, recalling the << of itself and recursiving till it crashes.
How should I proceed?
Thank you.
Easy! You need to invoke the version that takes an Abonne. So simply do that:
return out << static_cast<const Abonne&>(etudiant) << " ... and more" << endl;
// ^^^^^^^^^^^^^^^^^^^^^^^^^^
Related
I have created a constructor
Location(double xIn,double yIn,string placeIn,int timeIn)
: x(xIn),y(yIn) ...so on {
Say I want to print Location home(x,y,place,time); that's in the main().
How would I do so? I've been searching around and was told to use operator<<. How would I implement this?
UPDATE: After creating some get methods and I tried doing,can't exactly compile it because of the problem
ostream &operator<<(ostream & o, const Location & rhs){
o << rhs.getX() << "," << rhs.getY() << "," << rhs.getPlace() << "," << rhs.getTime();
return o; }
Here is the stencil for overloading operator<<:
class Any
{
public:
friend std::ostream& operator<<(std::ostream& output, const Any& a);
private:
int member;
};
std::ostream&
operator<<(std::ostream& output, const Any& a)
{
output << a.member;
return output;
}
This one possible stencil, there are other possibilities. Search the internet for "c++ stream insertion operator overload example" for other implementations.
class Sale
{
void Display();
I can call it in another file, let say register.cpp file as
something.Display();
How can I call Display() if were a friend operator<< function?
class Sale
{
friend ostream& operator<<(ostream& os, const Song& s);
here is sale.cpp with ostream definition
ostream& operator<<(ostream & os, const Song& s)
{ //title, artist, category, and size are private member data of sale.h
os << s.title << "\t" << s.artist
<< "\t" << s.category << "\t" << s.size;
}
Display() definition is hard-coded to cout
My assignment also requires me to call the operator<< in register.cpp file
to do the same thing as Display()
How can I do it?
I'm working on a project for my school in C++
I have 2 class : Employe and Teacher.
Teacher derived from Employe and has overrides of his functions.
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.
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;
}
But, the function calls itself with the line "os << (pTeacher);" and does a loop that causes a stack overflow.
I want that the line "os << (pTeacher)" calls the operator<< of my class Employe and not of my class Teacher.
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;
}
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).
I also tried to use Employe::operator<< but operator<< is not a member of Employe so it doesn't work...
So, here is my question :
How can I do to use my operator<< of Employe in my operator<< of Teacher and print the right information (LevelAccess = 20 and not 5) ?
I was also thinking of "virtual" but our professor tells us that there is not need to use this word.
Thanks in advance :)
Here is a more complete code :
main.cpp:
Teacher Garry("Garry");
cout << Garry << endl;
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;
}
With this in Employe.h :
private:
static const unsigned int LevelAccess_ = 5;
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;
}
With this is Teacher.h :
static const unsigned int accesTeacher_ = 15;
What I'd do is the following: define only one
ostream& operator<<(ostream& os, const Employe& pEmploye)
{
return pEmploye.display(os);
}
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;
}
};
You can use a cast:
os << static_cast<const Employe &>(pTeacher);
The & is important.
To make the call to the member function call Teacher::getLevelAccess() from an Employe reference, you have to make that function virtual. (Do this in teacher.h). getClass() should be virtual also.
NB. You keep saying things like "Override of operator<< in Employe :" , however you do not have overloaded operator<< in Employe . You have a free function which takes Employe as argument.
I want to overload the << operator for my class Complex.
The prototype is the following:
void Complex::operator << (ostream &out)
{
out << "The number is: (" << re <<", " << im <<")."<<endl;
}
It works, but I have to call it like this: object << cout for the standart output.
What can I do to make it work backwards, like cout << object?
I know that the 'this' pointer is by default the first parameter sent to the method so that's why the binary operator can work only obj << ostream. I overloaded it as a global function and there were no problems.
Is there a way to overload the << operator as a method and to call it ostream << obj?
I would just use the usual C++ pattern of free function. You can make it friend to your Complex class if you want to make Complex class's private data members visible to it, but usually a complex number class would expose public getters for real part and imaginary coefficient.
class Complex
{
....
friend std::ostream& operator<<(std::ostream &out, const Complex& c);
private:
double re;
double im;
};
inline std::ostream& operator<<(std::ostream &out, const Complex& c)
{
out << "The number is: (" << c.re << ", " << c.im << ").\n";
return out;
}
You could write a free stand operator<< function, try:
std::ostream& operator<< (std::ostream &out, const Complex& cm)
{
out << "The number is: (" << cm.re <<", " << cm.im <<")." << std::endl;
return out;
}
You can define a global function:
void operator << (ostream& out, const Complex& complex)
{
complex.operator<<(out);
}
Which operator do I have to overload if I want to use sth like this?
MyClass C;
cout<< C;
The output of my class would be string.
if you've to overload operator<< as:
std::ostream& operator<<(std::ostream& out, const MyClass & obj)
{
//use out to print members of obj, or whatever you want to print
return out;
}
If this function needs to access private members of MyClass, then you've to make it friend of MyClass, or alternatively, you can delegate the work to some public function of the class.
For example, suppose you've a point class defined as:
struct point
{
double x;
double y;
double z;
};
Then you can overload operator<< as:
std::ostream& operator<<(std::ostream& out, const point & pt)
{
out << "{" << pt.x <<"," << pt.y <<"," << pt.z << "}";
return out;
}
And you can use it as:
point p1 = {10,20,30};
std::cout << p1 << std::endl;
Output:
{10,20,30}
Online demo : http://ideone.com/zjcYd
Hope that helps.
The stream operator: <<
You should declare it as a friend of your class:
class MyClass
{
//class declaration
//....
friend std::ostream& operator<<(std::ostream& out, const MyClass& mc);
}
std::ostream& operator<<(std::ostream& out, const MyClass& mc)
{
//logic here
}
You should implement operator<< as a free function.