Question

(edited from original post to change “BaseMessage” to “const BaseMessage&”)

Hello All,
I’m very new to C++, so I hope you folks can help me “see the errors of my ways”.

I have a hierarchy of messages, and I’m trying to use an abstract base class to enforce
an interface. In particular, I want to force each derived message to provide an overloaded
<< operator.

When I try doing this with something like this:

class BaseMessage
{
public:

// some non-pure virtual function declarations
// some pure virtual function declarations

virtual ostream& operator<<(ostream& stream, const BaseMessage& objectArg) = 0;

}

the compiler complains that

“error: cannot declare parameter ‘objectArg’ to be of abstract type ‘BaseMessage’

I believe there are also “friend” issues involved here, but when I tried to declare it as:

virtual friend ostream& operator<<(ostream& stream, const BaseMessage objectArg) = 0;

the compiler added an addition error

“error: virtual functions cannot be friends”

Is there a way to ensure that all of my derived (message) classes provide an “<<” ostream operator?

Thanks Much,

Steve

Answer 1

The common convention for this is to have a friend output operator at the base level and have it call private virtual function:

class Base
{
public:

    /// don't forget this
    virtual ~Base();

    /// std stream interface
    friend std::ostream& operator<<( std::ostream& out, const Base& b )
    {
        b.Print( out );
        return out;
    }

private:

    /// derivation interface
    virtual void Print( std::ostream& ) const =0;
};