CODEWITHSUNDEEP
c++friend
starcorn
starcorn

Reputation: 8551

friend with class but can't access private members

Friend functions should be able to access a class private members right? So what have I done wrong here? I've included my .h file with the operator<< I intent to befriend with the class.

#include <iostream>

using namespace std;
class fun
{
private:
    int a;
    int b;
    int c;


public:
    fun(int a, int b);
    void my_swap();
    int a_func();
    void print();

    friend ostream& operator<<(ostream& out, const fun& fun);
};

ostream& operator<<(ostream& out, fun& fun)
{
    out << "a= " << fun.a << ", b= " << fun.b << std::endl;

    return out;
}

Upvotes: 10

Views: 12417

Answers (3)

fredoverflow
fredoverflow

Reputation: 263360

You can avoid these kinds of errors by writing the friend function definition inside the class definition:

class fun
{
    //...

    friend ostream& operator<<(ostream& out, const fun& f)
    {
        out << "a= " << f.a << ", b= " << f.b << std::endl;
        return out;
    }
};

The downside is that every call to operator<< is inlined, which might lead to code bloat.

(Also note that the parameter cannot be called fun because that name already denotes a type.)

Upvotes: 0

Puppy
Puppy

Reputation: 147056

The signatures don't match. Your non-member function takes fun& fun, the friend declared on takes const fun& fun.

Upvotes: 6

Brian Hooper
Brian Hooper

Reputation: 22084

In here...

ostream& operator<<(ostream& out, fun& fun)
{
    out << "a= " << fun.a << ", b= " << fun.b << std::endl;

    return out;
}

you need

ostream& operator<<(ostream& out, const fun& fun)
{
    out << "a= " << fun.a << ", b= " << fun.b << std::endl;

    return out;
}

(I've been bitten on the bum by this numerous times; the definition of your operator overload doesn't quite match the declaration, so it is thought to be a different function.)

Upvotes: 13

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