virtual class with friend functions
 

this is a similar example to what i have. it's simplified but it still get's the same error.

i need to be able to use << operator to print out all information from class A. i have no idea how to do that. i tried casting it as an A object but then it starts telling me that i can do that with virtual functions. is there any way to go around this?

#include <iostream>

#include <string>

using namespace std;



class A {

public:

        string name;

        A(string sName) {

                name = sName;

        }

        virtual void mission();

        string getName() {

                return name;

        }

        friend ostream& operator << (ostream& out, A& oA) {

                out << oA.getName() << endl;

                return out;

        }

};



class B : public A {

public:

        B() : A("bob") {

                name = "bob";

        }

};



int main() {



        B DUDE;



        cout << DUDE;

        // this doesnt work either

        cout << (A)DUDE;



        return 0;

}

Re: virtual class with friend functions
 

comment out

virtual void mission();

and try

and give different names for base class and derived class 'name' variable to see the difference.

Re: virtual class with friend functions
 

What's a problem?

class A {

public:

    A() {}

    explicit A(const std::string& n):name(n) {}

    const std::string& getName() const {

        return name;

    }

protected:

    std::string name;

};



inline

std::ostream& operator<<(std::ostream& os,const A& a) {

    os << a.getName();

    return os;

}



class B: public A {

public:

    B():A() {}

    explicit B(const std::string& n):A(n) {}

};



inline 

std::ostream& operator<<(std::ostream& os,const B& b) {

    return operator<<(os,static_cast<const A&>(b));

};



void abtest()

{

    B dude("mbayabo");

    std::cout << dude << std::endl;

}

Re: virtual class with friend functions
 

mayabo's original code (other than the line

cout << (A)DUDE;

) should actually work. The line

name="bob";

in B's constructor is not required (and A::name can be made private).

The compiler is able to do the conversion "derived class to public base" (i.e. B to A) implicitly.

If you're getting errors, then either there is some code in play not being shown or you have a buggy compiler (really old versions of VC++ come to mind).

It would probably be better to make getName() a const method (optionally virtual) and for the operator<<() to have its second argument a const reference.

And

"using namespace std;"

before a class definition: wash your mouth out with soap, particularly if you do that in a header file.....