CODEWITHSUNDEEP
c++operator-overloading
Scuba Steve
Scuba Steve

Reputation: 1648

friend function can't see private member variables

I'm trying to overload the << operator using a friend function, but it's not seeing the private member variables for some reason. Any ideas why this is happening would be quite helpful.

Here's the header file

    class Set
    {
private:
    struct Node
    {
        int val;
        Node* link;
    };

    Node *cons(int x, Node *p);
    Node *list;


public:
    Set() {list = NULL;}
    bool isEmpty() const;
    int size() const;
    bool member (int x) const;
    bool insert (int x);
    bool remove (int x);
    void print() const;
    const Set operator+(const Set &otherSet)const;
    const Set operator*(const Set &otherSet)const;
    Node* getSet()const{return list;}
    void setList(Node *list);
    friend ostream& operator <<(ostream& outputStream, const Set &set);
    };

Here's the function definition.

    ostream& operator <<(ostream& outputStream, const Set &set)
    {
              Node *p;
              p = list;
              outputStream << '{';
              while(p->link != NULL)
              {
                outputStream << p->val;
                p = p->link;
              }
              outputStream << '}';
              return outputStream;
    }

Upvotes: 2

Views: 2923

Answers (2)

Alok Save
Alok Save

Reputation: 206656

A simplistic representation of your code is:

class A
{
    struct MY
    {
        int a;
    };
    friend void doSomething(A);
};

void doSomething(A)
{
    MY *b;

}
int main()
{
    return 0;
}

The problem lies in: 

MY *b;

Your function cannot understand the type of MY since it is declared inside the class A. Hence the error:

In function 'void doSomething(A)':
Line 12: error: 'MY' was not declared in this scope
compilation terminated due to -Wfatal-errors.

In order to tell the function to find My inside A you need to use fully qualified name of the structure:

A::MY *b;

Once you do that the function knows exactly where to look for MY and hence can find it.
Working online sample.

Upvotes: 1

Sergey Kalinichenko
Sergey Kalinichenko

Reputation: 727137

The problem is not with the accessibility of Node but rather with its scope: the unqualified type name does not become in scope through friendship - you should use Set::Node instead.

Same goes for the list variable: it should be set.list.

With these two changes in place, your code compiles fine on ideone.

Upvotes: 5

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