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Home/ Questions/Q 8434887
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Editorial Team
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Editorial Team
Asked: June 10, 20262026-06-10T06:46:50+00:00 2026-06-10T06:46:50+00:00

I’m in the proccess of learning the language and this is a noob doubt.

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I’m in the proccess of learning the language and this is a noob doubt.

Is it possible to use a virtual friend function? I don’t know if it’s possible, I didn’t even test it but it could be useful in some situations. For example, for the overloaded operator<<().

DerivedClass dc;
BaseClass &rbc = dc;
cout << rbc;

My guess is it’s possible, but I’m not sure since a friend function is not implemented in the class design, and theoretically is not part of it (though in this example, conceptually it makes sense that operator<<() should be a method, but due to syntax limitations it’s not possible to implement it as one).

EDIT: my concern is related with this example:

BaseClass bc;
DerivedClass dc;
BaseClass *pArr[2];
pArr[1] = bc;
pArr[2] = dc;
for (int i = 0; i < 2; i++)
    cout << pArr[i];

in this array of mixed objects, I want the correct operator<<() called for each one.

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  1. Editorial Team
    Editorial Team
    2026-06-10T06:46:51+00:00Added an answer on June 10, 2026 at 6:46 am

    Nope, friend virtual functions doesn’t make sense at all.

    friend functions are such, that are not methods (a.k.a. member functions) and have the right to access private/protected members of a class.

    virtual functions can only be member functions. You can’t have virtual non-member function.


    You can make the operator<< take a reference to a base class and then call some virtual member function. This way, you can make the operator<< “almost virtual” 🙂


    For example

    class A
    {
    public:
        virtual void f() const { std::cout << "base"; }
    };
    class B: public A
    {
    public:
        virtual void f() const { std::cout << "derived"; }
    };
    
    std::ostream& operator<<(std::ostream& os, const A& a )
    {
         a.f();
         return os;
    }
    
    int main()
    {
        B b;
        std::cout << b << std::endl;
    
        return 0;
    }
    

    will print derived.

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