I have an assignment question for my object orientated programming course that asks for the creation of a class.
I am a little stuck on this part of the questions
The class must, however, be readable using the >> operator and writable using the << operator. Do not use a friend function to overload the operators. Instead create suitable read and write methods and then overload the operators using a non-friend function.
class MyClass
{
public:
void ReadFrom(std::istream &is)
{
// read values from 'is' as needed...
}
void WriteTo(std::ostream &os) const
{
// write values to 'os' as needed...
}
};
std::istream& operator>>(std::istream &is, MyClass &cls)
{
cls.ReadFrom(is);
return st;
}
std::ostream& operator<<(std::ostream &os, const MyClass &cls)
{
cls.WriteTo(os);
return os;
}
Related
Reading some documentation online I found that istream class was part of C++ long before the string class was added. So the istream design recognizes basic C++ types such as double and int, but it is ignorant of the string type. Therefore, there are istream class methods for processing double and int and other basic types, but there are no istream class methods for processing string objects.
My question is if there are no istream class methods for processing string objects, why this program works and how ?
#include <iostream>
int main(void)
{
std::string str;
std::cin >> str;
std::cout << str << std::endl;
return 0;
}
This is possible with the use operator overloading. As shown in the below example, you can create your own class and overload operator>> and operator<<.
#include <iostream>
class Number
{
//overload operator<< so that we can use std::cout<<
friend std::ostream& operator<<(std::ostream &os, const Number& num);
//overload operator>> so that we can use std::cin>>
friend std::istream &operator>>(std::istream &is, Number &obj);
int m_value = 0;
public:
Number(int value = 0);
};
Number::Number(int value): m_value(value)
{
}
std::ostream& operator<<(std::ostream &os, const Number& num)
{
os << num.m_value;
return os;
}
std::istream &operator>>(std::istream &is, Number &obj)
{
is >> obj.m_value;
if (is) // check that the inputs succeeded
{
;//do something
}
else
{
obj = Number(); // input failed: give the object the default state
}
return is;
}
int main()
{
Number a{ 10 };
std::cout << a << std::endl; //this will print 10
std::cin >> a; //this will take input from user
std::cout << a << std::endl; //this will print whatever number (m_value) the user entered above
return 0;
}
By overloading operator>> and operator<<, this allows us to write std::cin >> a and std::cout << a in the above program.
Similar to the Number class shown above, the std::string class also makes use of operator overloading. In particular, std::string overloads operator>> and operator<<, allowing us to write std::cin >> str and std::cout << str, as you did in your example.
Because std::string overload the >> and << operator to return the type std::istream and std::ostream
How they overload it, you can look in this link that Mat gives.
You can create your own class and overload operators, too. Here is an example:
class MyClass
{
int numberOne;
double numberTwo;
public:
friend std::ostream& operator<<(std::ostream &out, const MyClass& myClass);
friend std::istream& operator>> (std::istream& in, MyClass& myClass);
};
// Since operator<< is a friend of the MyClass class, we can access MyClass's members directly.
std::ostream& operator<<(std::ostream &out, const MyClass& myClass)
{
out << myClass.numberOne << ' ' << myClass.numberTwo;
return os;
}
// Since operator>> is a friend of the MyClass class, we can access MyClass's members directly.
std::istream& operator>> (std::istream& in, MyClass& myClass)
{
in >> myClass.numberOne;
in >> myClass.numberTwo;
return in;
}
int main()
{
MyClass myClass;
std::cin >> myClass;
std::cout << myClass;
}
Because of operator overloading.
In your case, including <iostream> will include <string>, a specialization of std::basic_string. And std::basic_string has the non-member functions operator<< and operator>> defined.
Similarly, you can also overload operator<< and operator>> for your own custom types.
That's basically the question, is there a "right" way to implement operator<< ?
Reading this I can see that something like:
friend bool operator<<(obj const& lhs, obj const& rhs);
is preferred to something like
ostream& operator<<(obj const& rhs);
But I can't quite see why should I use one or the other.
My personal case is:
friend ostream & operator<<(ostream &os, const Paragraph& p) {
return os << p.to_str();
}
But I could probably do:
ostream & operator<<(ostream &os) {
return os << paragraph;
}
What rationale should I base this decision on?
Note:
Paragraph::to_str = (return paragraph)
where paragraph's a string.
The problem here is in your interpretation of the article you link.
Equality
This article is about somebody that is having problems correctly defining the bool relationship operators.
The operator:
Equality == and !=
Relationship < > <= >=
These operators should return a bool as they are comparing two objects of the same type. It is usually easiest to define these operators as part of the class. This is because a class is automatically a friend of itself so objects of type Paragraph can examine each other (even each others private members).
There is an argument for making these free standing functions as this lets auto conversion convert both sides if they are not the same type, while member functions only allow the rhs to be auto converted. I find this a paper man argument as you don't really want auto conversion happening in the first place (usually). But if this is something you want (I don't recommend it) then making the comparators free standing can be advantageous.
Streaming
The stream operators:
operator << output
operator >> input
When you use these as stream operators (rather than binary shift) the first parameter is a stream. Since you do not have access to the stream object (its not yours to modify) these can not be member operators they have to be external to the class. Thus they must either be friends of the class or have access to a public method that will do the streaming for you.
It is also traditional for these objects to return a reference to a stream object so you can chain stream operations together.
#include <iostream>
class Paragraph
{
public:
explicit Paragraph(std::string const& init)
:m_para(init)
{}
std::string const& to_str() const
{
return m_para;
}
bool operator==(Paragraph const& rhs) const
{
return m_para == rhs.m_para;
}
bool operator!=(Paragraph const& rhs) const
{
// Define != operator in terms of the == operator
return !(this->operator==(rhs));
}
bool operator<(Paragraph const& rhs) const
{
return m_para < rhs.m_para;
}
private:
friend std::ostream & operator<<(std::ostream &os, const Paragraph& p);
std::string m_para;
};
std::ostream & operator<<(std::ostream &os, const Paragraph& p)
{
return os << p.to_str();
}
int main()
{
Paragraph p("Plop");
Paragraph q(p);
std::cout << p << std::endl << (p == q) << std::endl;
}
You can not do it as a member function, because the implicit this parameter is the left hand side of the <<-operator. (Hence, you would need to add it as a member function to the ostream-class. Not good :)
Could you do it as a free function without friending it? That's what I prefer, because it makes it clear that this is an integration with ostream, and not a core functionality of your class.
If possible, as non-member and non-friend functions.
As described by Herb Sutter and Scott Meyers, prefer non-friend non-member functions to member functions, to help increase encapsulation.
In some cases, like C++ streams, you won't have the choice and must use non-member functions.
But still, it does not mean you have to make these functions friends of your classes: These functions can still acess your class through your class accessors. If you succeed in writting those functions this way, then you won.
About operator << and >> prototypes
I believe the examples you gave in your question are wrong. For example;
ostream & operator<<(ostream &os) {
return os << paragraph;
}
I can't even start to think how this method could work in a stream.
Here are the two ways to implement the << and >> operators.
Let's say you want to use a stream-like object of type T.
And that you want to extract/insert from/into T the relevant data of your object of type Paragraph.
Generic operator << and >> function prototypes
The first being as functions:
// T << Paragraph
T & operator << (T & p_oOutputStream, const Paragraph & p_oParagraph)
{
// do the insertion of p_oParagraph
return p_oOutputStream ;
}
// T >> Paragraph
T & operator >> (T & p_oInputStream, const Paragraph & p_oParagraph)
{
// do the extraction of p_oParagraph
return p_oInputStream ;
}
Generic operator << and >> method prototypes
The second being as methods:
// T << Paragraph
T & T::operator << (const Paragraph & p_oParagraph)
{
// do the insertion of p_oParagraph
return *this ;
}
// T >> Paragraph
T & T::operator >> (const Paragraph & p_oParagraph)
{
// do the extraction of p_oParagraph
return *this ;
}
Note that to use this notation, you must extend T's class declaration. For STL objects, this is not possible (you are not supposed to modify them...).
And what if T is a C++ stream?
Here are the prototypes of the same << and >> operators for C++ streams.
For generic basic_istream and basic_ostream
Note that is case of streams, as you can't modify the C++ stream, you must implement the functions. Which means something like:
// OUTPUT << Paragraph
template <typename charT, typename traits>
std::basic_ostream<charT,traits> & operator << (std::basic_ostream<charT,traits> & p_oOutputStream, const Paragraph & p_oParagraph)
{
// do the insertion of p_oParagraph
return p_oOutputStream ;
}
// INPUT >> Paragraph
template <typename charT, typename traits>
std::basic_istream<charT,traits> & operator >> (std::basic_istream<charT,traits> & p_oInputStream, const CMyObject & p_oParagraph)
{
// do the extract of p_oParagraph
return p_oInputStream ;
}
For char istream and ostream
The following code will work only for char-based streams.
// OUTPUT << A
std::ostream & operator << (std::ostream & p_oOutputStream, const Paragraph & p_oParagraph)
{
// do the insertion of p_oParagraph
return p_oOutputStream ;
}
// INPUT >> A
std::istream & operator >> (std::istream & p_oInputStream, const Paragraph & p_oParagraph)
{
// do the extract of p_oParagraph
return p_oInputStream ;
}
Rhys Ulerich commented about the fact the char-based code is but a "specialization" of the generic code above it. Of course, Rhys is right: I don't recommend the use of the char-based example. It is only given here because it's simpler to read. As it is only viable if you only work with char-based streams, you should avoid it on platforms where wchar_t code is common (i.e. on Windows).
Hope this will help.
It should be implemented as a free, non-friend functions, especially if, like most things these days, the output is mainly used for diagnostics and logging. Add const accessors for all the things that need to go into the output, and then have the outputter just call those and do formatting.
I've actually taken to collecting all of these ostream output free functions in an "ostreamhelpers" header and implementation file, it keeps that secondary functionality far away from the real purpose of the classes.
The signature:
bool operator<<(const obj&, const obj&);
Seems rather suspect, this does not fit the stream convention nor the bitwise convention so it looks like a case of operator overloading abuse, operator < should return bool but operator << should probably return something else.
If you meant so say:
ostream& operator<<(ostream&, const obj&);
Then since you can't add functions to ostream by necessity the function must be a free function, whether it a friend or not depends on what it has to access (if it doesn't need to access private or protected members there's no need to make it friend).
Just for completion sake, I would like to add that you indeed can create an operator ostream& operator << (ostream& os) inside a class and it can work. From what I know it's not a good idea to use it, because it's very convoluted and unintuitive.
Let's assume we have this code:
#include <iostream>
#include <string>
using namespace std;
struct Widget
{
string name;
Widget(string _name) : name(_name) {}
ostream& operator << (ostream& os)
{
return os << name;
}
};
int main()
{
Widget w1("w1");
Widget w2("w2");
// These two won't work
{
// Error: operand types are std::ostream << std::ostream
// cout << w1.operator<<(cout) << '\n';
// Error: operand types are std::ostream << Widget
// cout << w1 << '\n';
}
// However these two work
{
w1 << cout << '\n';
// Call to w1.operator<<(cout) returns a reference to ostream&
w2 << w1.operator<<(cout) << '\n';
}
return 0;
}
So to sum it up - you can do it, but you most probably shouldn't :)
friend operator = equal rights as class
friend std::ostream& operator<<(std::ostream& os, const Object& object) {
os << object._atribute1 << " " << object._atribute2 << " " << atribute._atribute3 << std::endl;
return os;
}
operator<< implemented as a friend function:
#include <iostream>
#include <string>
using namespace std;
class Samp
{
public:
int ID;
string strName;
friend std::ostream& operator<<(std::ostream &os, const Samp& obj);
};
std::ostream& operator<<(std::ostream &os, const Samp& obj)
{
os << obj.ID<< “ ” << obj.strName;
return os;
}
int main()
{
Samp obj, obj1;
obj.ID = 100;
obj.strName = "Hello";
obj1=obj;
cout << obj <<endl<< obj1;
}
OUTPUT:
100 Hello
100 Hello
This can be a friend function only because the object is on the right hand side of operator<< and argument cout is on the left hand side. So this can't be a member function to the class, it can only be a friend function.
I'm in a programming class and need overloading explained to me. Simple question so hopefully I'll get an answer pretty quick. My understanding is that overloading an operator allows it to be used on a class. If that is true, then how would I overload >> to work with a class? I'm working on a small program to test out this idea and i'll post it here
#include <iostream>
#include <cstdlib>
#include "data.h"
using namespace std;
int main()
{
data obj;
cout << "What is the Number?" << endl;
cin >> obj;
system("pause");
return 0;
}
class data
{
public:
data operator >> (int);
private:
};
This page tells you mostly everything you need to know about operator overloading.
In short, nearly every operator in C++ can be overloaded for user-defined types. Some operators, like +, -, or >> must be defined outside of a class since they are free-standing, whereas others like copy assignment (=), must be defined within.
For your case, overloading the >> operator can be done in the following manner:
istream& operator>>(istream& in, data& d){
// Code here
return in;
}
Where it says "Code here", place the code you need to read into the data object.
For example, let us pretend that we were reading into a Point object with an x coordinate and a y coordinate. It is formatted in the stream like so: "(x,y)". The operator overload might look like this:
istream& operator>>(istream& in, Point& p){
char c;
in >> c;
if(c != '('){
in.clear(ios_base::failbit);
return in;
}
in >> p.x >> c >> p.y >> c;
return in;
}
This is just an example with minimal format checking, but hopefully it is enough to get you started.
Note that if members in your class are private, then you should friend the istream operator in the class definition:
class data{
...
public:
friend istream& operator>>(istream&, data&);
}
case1: no need to access private data
data.h.
class data {
public:
int i;
};
std::ostream& operator>> (std::istream&, data&); // better make operator >>
// a nonmember function
// if it doesn't need access
// to private data
data.cpp
#include "data.h"
std::istream& operator>> (std::istream& is, data& d) {
is>>d.i; // here we do some logic, we do what it means to do >> on
return is; // an instance of your data class and return reference to istream
}
case2: there is a need to access private data
data.h.
class data {
private:
int i;
friend std::ostream& operator>> (std::istream&, data&);
};
data.cpp
#include "data.h"
std::istream& operator>> (std::istream& is, data& d) {
is>>d.i; // here we do some logic, we do what it means to do >> on
return is; // an instance of your data class and return reference to istream
}
If you want to bolster your understanding of what operator overloading is, consider that essentially all operators on objects (such as "+", "++", "==", "!=", etc) are member functions.
Challenge yourself to recognize Obj a, b; a = b; as Obj a; Obj b; a.operator=(b);.
Overloading is purely providing a non-default implementation.
Here is a [terrible] overload of the cast-to-const-char* operator:
class BadWolf {
const char* m_text;
public:
BadWolf(const char* text) : m_text(text) {}
// if you really want the original text and ask for it with this function...
const char* tardisTranslation() const { return m_text; }
// but if you try to use me as a const char*...
operator const char* () const { return "bad wolf"; }
};
int main(int argc, const char* argv[]) {
BadWolf message("hello, sweetie");
std::cout << "Message reads: " << (const char*)message << std::endl;
std::cout << "Tardis translation: " << message.tardisTranslaction() << std::endl;
return 0;
}
I think this is what you want.
class data
{
friend istream& operator>>( istream&, data& );
private:
int data;
};
istream& operator>>( istream& in, data& d )
{
return in >> d.data;
}
I'm trying to write a class that overloads the insertion operator but in my header file I get the error.
Overloaded 'operator<<' must be a binary operator (has 3 parameters)
Here is my code:
.h file
ostream & operator<<(ostream & os, Domino dom);
.cpp file
ostream & operator<< (ostream & os, Domino dom) {
return os << dom.toString();
}
I'm following a text book and this is what they use as an example but its not working for me.. Any suggestions?
You probably put your operator<< inside a class declaration. That means it takes an extra hidden parameter (the this parameter). You need to put it outside of any class declaration.
The insertion operator (<<) can be used as a member function or a friend function.
operator << used as a member function
ostream& operator<<(ostream& os);
This function should be invoked as :
dom << cout;
In general if you are using the operator as a member function, the left hand side of the operator should be an object. Then this object is implicitly passed as an argument to the member function. But the invocation confuses the user and it does not look nice.
operator << used as a friend function
friend ostream& operator<<(ostream& os, const Domino& obj);
This function should be invoked as :
cout << dom;
In this case the object dom is explicitly passed as a reference. This invocation is more traditional and user can easily understand the meaning of the code.
/*insertion and extraction overloading*/
#include<iostream>
using namespace std;
class complex
{
int real,imag;
public:
complex()
{
real=0;imag=0;
}
complex(int real,int imag)
{
this->real=real;
this->imag=imag;
}
void setreal(int real)
{
this->real=real;
}
int getreal()
{
return real;
}
void setimag(int imag)
{
this->imag=imag;
}
int getimag()
{
return imag;
}
void display()
{
cout<<real<<"+"<<imag<<"i"<<endl;
}
};//end of complex class
istream & operator >>(istream & in,complex &c)
{
int temp;
in>>temp;
c.setreal(temp);
in>>temp;
c.setimag(temp);
return in;
}
ostream &operator <<(ostream &out,complex &c)
{
out<<c.getreal()<<c.getimag()<<endl;
return out;
}
int main()
{
complex c1;
cin>>c1;
// c1.display();
cout<<c1;
//c1.display();
return 0;
}
I'm a bit confused about how to overload the stream operators for my class in C++, since it seems they are functions on the stream classes, not on my class. What's the normal way to do this? At the moment, for the "get from" operator, I have a definition
istream& operator>>(istream& is, Thing& thing) { // etc...
which works. It's not mentioned in the definition of the Thing class. I want it to be able to access members of my Thing class in its implementation - how do I do this?
Your implementation is fine. The only additional step you need to perform is to declare your operator as a friend in Thing:
class Thing {
public:
friend istream& operator>>(istream&, Thing&);
...
}
The other answers are right. In case it helps you, here's a code example (source):
class MyClass {
int x, y;
public:
MyClass(int i, int j) {
x = i;
y = j;
}
friend ostream &operator<<(ostream &stream, MyClass ob);
friend istream &operator>>(istream &stream, MyClass &ob);
};
ostream &operator<<(ostream &stream, MyClass ob)
{
stream << ob.x << ' ' << ob.y << '\n';
return stream;
}
istream &operator>>(istream &stream, MyClass &ob)
{
stream >> ob.x >> ob.y;
return stream;
}
You make your operator>> a friend of the Thing class.
There are several approaches and the right one really depends on what the class does.
Quite often it makes sense to have public API that allows reading the information, in which case the streaming operators do not need to access privates.
Probably the most popular approach is declaring the streaming functions friends of the class.
My favorite is providing a public Boost.Serialization style template function that can be used for streaming either way, as well as for other things.