Im really unsure how to call the function:
friend ostream& operator<<(ostream& out, stack::myItem& theItem);
that is public to my stack object:
class stack
{
public:
stack(int capacity);
~stack(void);
void method1();
...
private:
struct myItem
{
int item;
};
...
public:
friend ostream& operator<<(ostream& out, stack& s);
friend ostream& operator<<(ostream& out, stack::myItem& theItem);
};
It's no different than using stream operator << for any other type (it is called operator overloading for a reason).
However, outputting should not modify an object, hence you really should pass it by const reference (otherwise calls with temporaries would fail to compile).
friend ostream& operator<<(ostream& out, const stack& s);
friend ostream& operator<<(ostream& out, const stack::myItem& theItem);
This operator is a classic binary operator.
// Say I have an operator declared like this:
return_type operator#(left_type lhs, right_type rhs);
// Then the invocation is done this way:
left_type L;
right_type R;
return_type result = L # R;
In the case of the streaming operator, it is a bit special since the left hand argument and the return type actually have the same type (and indeed, will refer to the same object, albeit at different times). This has been done to allow chaining.
// Chaining
std::cout << "<Output> " << 1 << std::endl;
// Which can be analyzed like such
operator<<(
operator<<(
operator<<(
std::cout ,
"<Output> "
),
1
),
std::endl
);
As you can see, the syntax merely allows a convenient invocation. One might note that the order is very well defined, it is a strict left to right evaluation.
So with your object, it would become:
stack s;
std::cout << s << std::endl;
Just like that!
Call it from where? As it's coded only the class knows about the private struct. No code external to the class could use that method since it couldn't create an instance of the struct. Marking it as friend doesn't do you much good.
Related
I have the following class template which will accept both primitives and object. However like this I can only print primitives. How can I make it function using both primitives and objects? Thanks
template<class T>
class A
{
private:
vector <T> l;
public:
void print() const
{
for (int i=0;i<.size();i++)
{
cout<<l[i]<<endl; //error here
}
}
};
The reason why you can print primitives is that <iostream> provides overloads for operator<< for them.
To let your template print your classes in the same way, you need to define your own implementation of the operator:
// This implementation puts operator << outside your class.
// Mark it "friend" in MyClass if it needs access to private members of MyClass.
ostream& operator<<(ostream& ostr, const MyClass& myClass) {
// Do the printing based on the members of your class
ostr << myClass.member1 << ":" << myClass.member2;
return ostr;
}
The compiler will detect this operator during template expansion, and use it for printing when you do this:
cout<<l[i]<<endl;
You can put operator<< inside your class as well:
ostream &operator<<(ostream &os) {
ostr << member1 << ":" << member2;
}
I am assuming that here, you want to print an object instead of a variable belonging to a fundamental datatype.
For such cases, you can look at operator overloading in C++(more specifically overloading insertion operator).
For more information about overloading the insertion operator for an object, you can visit this URL
http://msdn.microsoft.com/en-us/library/1z2f6c2k.aspx
Given below is an example about how to go about it
ostream& operator<<(ostream& os, const Datatype& dt)
{
os << dt.a <<" " << dt.b;
return os;
}
Here Datatype is the name of the class and a and b are two private members of a and b which will be printed when you try to print the object.
However, to overload using this technique, do not forget to make this function as a friend function of the class(as given below) as the function requires access the to private members of the class.
friend ostream& operator<<(ostream& os, const Datatype& dt);
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 have a class Counter and I want to overload operator << to output the data member of Counter. I tried to make the ostream overloading a member function:
Counter{
public:
std::ostream& operator<<(std::ostream& outStream, const Counter& c);
private:
int count_;
};
std::ostream& Counter::operator<<(std::ostream& outStream, const Counter& c){
outStream << c.count_;
return outStream;
}
But the g++ compiler always outputs the same error:
‘std::ostream& Counter::operator<<(std::ostream&, const Counter&)’ must take exactly one argument
However, if I changed the overloading function to be a friend of the class, it worked all well, like this:
Counter{
public:
friend std::ostream& operator<<(std::ostream& outStream, const Counter& c);
private:
int count_;
};
std::ostream& operator<<(std::ostream& outStream, const Counter& c){
outStream << c.count_;
return outStream;
}
Does this mean that the the stream operator overloading cannot be a member function of a class?
Add a public query method that returns the value of count_, then it does not have to be a friend:
Counter{
public:
int count() const { return count_; }
private:
int count_;
};
std::ostream& operator<<(std::ostream& outStream, const Counter& c){
outStream << c.count();
return outStream;
}
If you put the ostream operator in the class itself then it will not work the way you expect it to. It would be a member function meaning to invoke it one would have to do this: c.operator<<("output") which is obviously not what you mean to do. For it to work as you expect an ostream operator it must be outside the class. You can do this by making it a friend or just put it outside of the class and use getters (accessors) to output the data.
It doesn't have to be a friend, but it can't be a member. Member operators only work when they are inside the class which corresponds to the left-hand operand.
Unfortunately the useful overloads for the streaming output operators ( << ) cannot be class members, because the ostream& must be on the left in use and declaration. They do not need to be friends of the class you wish to stream unless they need access to protected or private members. This means that if you can implement a streaming operator using just public functions such as observers/accessors without declaring it a friend.
In your first Counter class you are declaring a member function of the class that does not seem valid. In the second example of the Counter class you are stating that your operator overload for << , which seems valid, has access to the private members. In the second example the function must still be declared outside the class.
Wikipedia Operators in C and C++ has a good list of possible operator overloads, including the in class << overloads even though they are not very useful. The in class overloads must be called backwards CounterInstance << cout; which is counterintuitive.
hey, i got something that i cannot understand ,there are two types of solutions for overloading this operator 1 is including the friend at the start of the method and the other 1 goes without the friend.
i would very much like if some1 explain whats the difference between them advantages / disadvantages.
for example overloading the operator << in class rational:
class Rational:
{
private: int m_t,m_b;
...
friend ostream& operator<<(ostream& out,const Rational& r) // option 1
{ return out << r.m_t << "/" <<r.m_b;} // continue of option 1
ostream& operator<<(ostream& out,const Rational& r){return r.print();} // option 2
virtual ostream& print(ostream& out) const // continue of option 2
{ //
return out<<m_t << "/" << m_b;
} //
};
i was told that the second option isnt correct , if some1 can correct me about it i would much appriciate it.
thanks in advance.
The short answer: Option #2 actually isn't an option, but a syntax error, because it tries to define a binary operator as a member passing two operands.
The somewhat longer answer: If you make the second operand a free function (not a member of the class), this will work. Which one is preferable depends on the circumstances and your preferences. For starters: The disadvantage of the first is that it allows operator<< to access everything in Rational (including private helper functions), while the disadvantage of the second is that you introduce a function to the class' public API that nobody needs.
operator<< (for ostream) needs to be a free function (since the left-hand argument is a stream, not your class).
The friend keyword makes it a free function (a free function that has access to the private members).
However, if this functionality can be implemented in terms of the public interface, it is better to do so and just use a non-friend free function.
class Rational:
{
private: int m_t,m_b;
public:
...
virtual ostream& print(ostream& out) const
{
return out<<m_t << "/" << m_b;
}
};
ostream& operator<<(ostream& out,const Rational& r)
{
return r.print(out);
}
Consider a function that should output the num and den of Rational:
ostream& operator<<(ostream& out, const Rational& r)
{
return out;
}
Unfortunately, this is just a global function. Like any other global function, it cannot access the private members of Rational. To make it work with Rational objects, you need to make it friend of Rational:
class Rational
{
private: int m_t,m_b;
// ...
friend ostream& operator<<(ostream& out, const Rational& r);
};
ostream& operator<<(ostream& out, const Rational& r)
{
out << r.m_t << "/" <<r.m_b;
return out;
}
The friend ostream& operator<<(ostream& out, const Rational& r); inside Rational class indicates that ostream& operator<<(ostream& out, const Rational& r) function can directly use Rational's private members.
Now when you write:
Rational r(1, 2); // Say, it sets num and den
cout << r;
the following function call is made:
operator<<(cout, r);
Can you write operator<< as a member function of Rational? That's simply not possible because of the above conversion where cout has to be first parameter. If you make operator<< as a member of Rational:
class Rational
{
private: int m_t,m_b;
// ...
public:
ostream& operator<<(ostream& out) const
{
out << r.m_t << "/" <<r.m_b;
return out;
}
};
you need to call it this way:
Rational r(1, 2);
r.operator<<(cout);
which is ugly.
In a project I'm working on, I have a Score class, defined below in score.h. I am trying to overload it so, when a << operation is performed on it, _points + " " + _name is printed.
Here's what I tried to do:
ostream & Score::operator<< (ostream & os, Score right)
{
os << right.getPoints() << " " << right.scoreGetName();
return os;
}
Here are the errors returned:
score.h(30) : error C2804: binary 'operator <<' has too many parameters
(This error appears 4 times, actually)
I managed to get it working by declaring the overload as a friend function:
friend ostream & operator<< (ostream & os, Score right);
And removing the Score:: from the function declaration in score.cpp (effectively not declaring it as a member).
Why does this work, yet the former piece of code doesn't?
Thanks for your time!
EDIT
I deleted all mentions to the overload on the header file... yet I get the following (and only) error. binary '<<' : no operator found which takes a right-hand operand of type 'Score' (or there is no acceptable conversion)
How come my test, in main(), can't find the appropriate overload? (it's not the includes, I checked)
Below is the full score.h
#ifndef SCORE_H_
#define SCORE_H_
#include <string>
#include <iostream>
#include <iostream>
using std::string;
using std::ostream;
class Score
{
public:
Score(string name);
Score();
virtual ~Score();
void addPoints(int n);
string scoreGetName() const;
int getPoints() const;
void scoreSetName(string name);
bool operator>(const Score right) const;
private:
string _name;
int _points;
};
#endif
Note: You might want to look at the operator overloading FAQ.
Binary operators can either be members of their left-hand argument's class or free functions. (Some operators, like assignment, must be members.) Since the stream operators' left-hand argument is a stream, stream operators either have to be members of the stream class or free functions. The canonical way to implement operator<< for any type is this:
std::ostream& operator<<(std::ostream& os, const T& obj)
{
// stream obj's data into os
return os;
}
Note that it is not a member function. Also note that it takes the object to stream per const reference. That's because you don't want to copy the object in order to stream it and you don't want the streaming to alter it either.
Sometimes you want to stream objects whose internals are not accessible through their class' public interface, so the operator can't get at them. Then you have two choices: Either put a public member into the class which does the streaming
class T {
public:
void stream_to(std::ostream&) const {os << obj.data_;}
private:
int data_;
};
and call that from the operator:
inline std::ostream& operator<<(std::ostream& os, const T& obj)
{
obj.stream_to(os);
return os;
}
or make the operator a friend
class T {
public:
friend std::ostream& operator<<(std::ostream&, const T&);
private:
int data_;
};
so that it can access the class' private parts:
inline std::ostream& operator<<(std::ostream& os, const T& obj)
{
os << obj.data_;
return os;
}
Let's say you wanted to write an operator overload for + so you could add two Score objects to each other, and another so you could add an int to a Score, and a third so you could add a Score to an int. The ones where a Score is the first parameter can be member functions of Score. But the one where an int is the first parameter can't become member functions of int, right? To help you with that, you're allowed to write them as free functions. That is what is happening with this << operator, you can't add a member function to ostream so you write a free function. That's what it means when you take away the Score:: part.
Now why does it have to be a friend? It doesn't. You're only calling public methods (getPoints and scoreGetName). You see lots of friend operators because they like to talk directly to the private variables. It's ok by me to do that, because they are written and maintained by the person maintaing the class. Just don't get the friend part muddled up with the member-function-vs-free-function part.
You're getting compilation errors when operator<< is a member function in the example because you're creating an operator<< that takes a Score as the first parameter (the object the method's being called on), and then giving it an extra parameter at the end.
When you're calling a binary operator that's declared as a member function, the left side of the expression is the object the method's being called on. e.g. a + b might works like this:
A a;
B b
a.operator+(b)
It's typically preferable to use non-member binary operators (and in some cases -- e.g. operator<<for ostream is the only way to do it. In that case, a + b might work like this:
A a;
B b
operator+(a, b);
Here's a full example showing both ways of doing it; main() will output '55' three times:
#include <iostream>
struct B
{
B(int b) : value(b) {}
int value;
};
struct A
{
A(int a) : value(a) {}
int value;
int operator+(const B& b)
{
return this->value + b.value;
}
};
int operator+(const A& a, const B& b)
{
return a.value + b.value;
}
int main(int argc, char** argv)
{
A a(22);
B b(33);
std::cout << a + b << std::endl;
std::cout << operator+(a, b) << std::endl;
std::cout << a.operator+(b) << std::endl;
return 0;
}