why is cout calling parent operator<< instead of child operator<<

Question

I have 3 classes, and I'd like each one to print out differently to the terminal, I have a node class that represents a vertex in a BDD graph, right now I'm trying to write code to do logical operations on the nodes.

The Node class is setup as such:

class Node {

    char name;

    public:
        Node() { name = '0'; }
        Node(char c) { name = c; }

        Node(const Node& n) { name = n.name; }

        friend ostream& operator<<(ostream& stream, const Node& n);
};

ostream& operator<<(ostream& stream, const Node& n) {
    return stream << "{ Node " << n.name << " }";
}

The operator classes are setup as such:

class Operation {
    public:

    Node result;

    friend std::ostream& operator<<(std::ostream& stream, const Operation& op);

    Operation() {}

    Operation(const Operation& op) { cout << "Copying " << *this << endl; }

    virtual ~Operation() { cout << "Destroying " << *this << endl; }

    virtual Node compute() { 
        cout << "Computing " << *this << endl; 
        result = Node('1');
        return result; 
    }
};

std::ostream& operator<<(std::ostream& stream, const Operation& op) { 
    return stream << "Operation { Unspecified }"; 
}


class UnaryOperation : public Operation {
    public:

    Node arg1;

    UnaryOperation(const Node& arg1) { this->arg1 = arg1; }

    UnaryOperation(const UnaryOperation& op) : Operation::Operation(op) {
        arg1 = op.arg1;
    }

    virtual ~UnaryOperation() {}

    friend ostream& operator<<(ostream& stream, const UnaryOperation& op);
};

ostream& operator<<(ostream& stream, const UnaryOperation& op) {
    return stream << "Operation { arg1: " << op.arg1 << " }";
}


class BinaryOperation : public UnaryOperation {
    public:

    Node arg2;

    BinaryOperation(const Node& arg1, const Node& arg2) : UnaryOperation(arg1) { 
        this->arg2 = arg2; 
    }

    BinaryOperation(const BinaryOperation& op) : UnaryOperation::UnaryOperation(op) {
        arg2 = op.arg2;
    }

    virtual ~BinaryOperation() {}

    friend ostream& operator<<(ostream& stream, const BinaryOperation& op);
};

ostream& operator<<(ostream& stream, const BinaryOperation& op) {
    return stream << "Operation { arg1: " << op.arg1 << ", arg2: " << op.arg2;
}

For debugging reasons, these messages need to print out as such, but when I run this

Node apply(Operation& op) {
    cout << "Performing apply operation on " << op << endl;
    op.compute();
    return op.result;
}


int main() {
    Node a('a'), b('b');
    UnaryOperation uop(a);
    BinaryOperation bop(a, b);
    cout << uop << endl;
    cout << bop << endl;
    apply(uop);
    apply(bop);
}

I get

Operation { arg1: { Node a } }
Operation { arg1: { Node a }, arg2: { Node b }
Performing apply operation on Operation { Unspecified }
Computing Operation { Unspecified }
Performing apply operation on Operation { Unspecified }
Computing Operation { Unspecified }
Destroying Operation { Unspecified }
Destroying Operation { Unspecified }

Needless to say, this is not very helpful for debugging.

Why is it doing this, and how do I fix it?

[Cannot duplicate](http://coliru.stacked-crooked.com/a/8841edb9a1528e37) — PaulMcKenzie, Apr 01 '20 at 16:44
`virtual ~BinaryOperation() : UnaryOperation::~UnaryOperation() {}` -- What is this supposed to be doing? — PaulMcKenzie, Apr 01 '20 at 16:54
That's an excellent question, hold on, I'm going to make sure this actually looks like the code I'm running — iggy12345, Apr 01 '20 at 16:59
This looks like you were trying to use the syntax for calling the constructor of a base class in a destructor. This doesn't compile and you don't need to that, the destructor of the base class will be called automatically. — Lukas-T, Apr 01 '20 at 17:07
@PaulMcKenzie Okay, this should compile, and it should show exactly what I'm trying to ask. — iggy12345, Apr 01 '20 at 17:34
@VladfromMoscow I apologize, I was working on getting that for you when you commented. — iggy12345, Apr 01 '20 at 17:35

score 1 · Accepted Answer · answered Apr 01 '20 at 17:45

The friend function is not virtual. It is selected according to the type of the second argument.

In this function

Node apply(Operation& op) {
    cout << "Performing apply operation on " << op << endl;
    op.compute();
    return op.result;
}

the type of the argument is Operation &. So in this statement

cout << "Performing apply operation on " << op << endl;

there is called the friend function for an object of the type Operation &.

You could make the friend function "virtual" the following way as it is shown in the demonstrative program below.

#include <iostream>

class Operation 
{
public: 
    virtual ~Operation() = default;

private:    
    virtual std::ostream & out( std::ostream &os = std::cout ) const
    {
        return os << "This is an Operation"; 
    }

    friend std::ostream& operator <<( std::ostream &stream, const Operation &op ) 
    { 
        return op.out( stream ); 
    }
};

class UnaryOperation : public Operation 
{
    std::ostream & out( std::ostream &os = std::cout ) const override
    {
        return os << "This is an Unary Operation";
    }

    friend std::ostream& operator <<( std::ostream &stream, const UnaryOperation &op ) 
    { 
        return op.out( stream ); 
    }
};

class BinaryOperation : public UnaryOperation 
{
    std::ostream & out( std::ostream &os = std::cout ) const override
    {
        return os << "This is a Binary Operation";
    }

    friend std::ostream& operator <<( std::ostream& stream, const BinaryOperation &op ) 
    { 
        return op.out( stream ); 
    }
};

void f( const Operation &op )
{
    std::cout << op << '\n';
}

int main() 
{
    BinaryOperation bop;

    f( bop );

    return 0;
}

Its output is

This is a Binary Operation

why is cout calling parent operator<< instead of child operator<<

1 Answers1