Overloading the stream insertion (<<) operator for a class

Question

It is often overloaded as a friend function of the class. Is there any way it can be overloaded as a member function?

Answer 1

Is there any way it can be overloaded as a member function?

No. The signature of the function prevents this option.

// Binary operator where the stream object is left of '<<' and the object
// instance is right of '<<'
std::ostream& operator<<(std::ostream& lhs, const Foo& rhs)

For illustration here's a binary operator+ as a free function:

class Foo
{
};

Foo operator+(const Foo& lhs, const Foo& rhs)
{
    // an appropriate implementation
}

int main()
{
    Foo f1;
    Foo f2;

    // Uses the free function
    Foo result = f1 + f2;

    return 0;
}

However implemented as a member function it looks like this:

class Foo
{
public:
    Foo operator+(const Foo& other) const
    {
        // an appropriate implementation
    }
};

int main()
{
    Foo f1;
    Foo f2;

    // The calling sequence is the same as before however 'f1' is logically
    // the same as 'lhs' in the free function implementation example (i.e.,
    // the first type in the binary operator must be the class type when
    // implemented as a member function)
    Foo result = f1 + f2;

    return 0;
}

Given the operator+ member function example it is easier to see why the std::ostream& operator<<(std::ostream& lhs, const Foo& rhs) free function cannot be converted to an equivalent member function. Because 'lhs' would need to be the same type as std::ostream but this is only possible when implementing that class whereas in this example the Foo class is being implemented.