CODEWITHSUNDEEP
c++operator-overloadingfriend-function
Hossein
Hossein

Reputation: 25944

Is there a way to overload extraction operator, inside a class and not as a friend of class?

I am trying to overload << operator, so that when I use my class object in std::cout, it prints out the data members which is supposed to get printed.

I know I have to define a friend functions signature inside of a class, and then outside of the class I define two friend functions doing the job, but I want to hand over the class, and don't want the user to do literally anything and I want the class take care of everything.

How can I achieve such a thing in C++?

Upvotes: 0

Views: 799

Answers (3)

Potatoswatter
Potatoswatter

Reputation: 137850

C++ is designed to allow non-member functions to do things that would require a member function in other languages. A friend function is very similar to a member function, in that it can be defined within a class scope. Also (and this can be confusing) if you declare a friend function only within a class scope, it can only be accessed by passing an object of that class to it.

struct printme {
    friend std::ostream &operator<< ( std::ostream &os, printme const & )
        { return os << "hello"; }

    friend void named_friend( printme const & ) {}
};

std::cout << printme(); // OK
named_friend( printme() ); // OK
void (*fn_ptr) = named_friend; /* function not found
           it can only be found by association with printme */

Although this subtle rule is odd at first, it reflects the idea that nonmember functions can also be encapsulated by the class.

For more info, read up on argument-dependent lookup (ADL).

Upvotes: 2

whitetiger
whitetiger

Reputation: 422

we usually uses << in the following way: 

cout<<myobject;

if you make operator<<() a member function of your class you have to use the following syntax to call operator<<()

myobject<<cout;

do you want it that way? if its ok for you , you can make it a member function .

if you want it like 

cout<<myobject

- you have to make operator<<() a friend function of your class

Upvotes: 0

Israel Unterman
Israel Unterman

Reputation: 13510

Here is an example:

#include <iostream>
#include <string>

using namespace std;

class A {
public:
    string pr() { return "Hi, I'm class A"; }
};

ostream &operator << (ostream &o, const A &a) {
    o << a.pr();
    return o;
}

int main() {
    A a;

    cout << a << endl;

    return 0;
}

However, from my experience, don't do that. Oveloading is academically nice, but it makes it diffult to search your application for its implementation, and even comprehend that << has been overloaded. I always use something of the form cout << myobject.tostring()

Upvotes: -1

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