Hi all,

when I write a subclass, can I inherit its superclass' << operator?

As an example say we have two classes, Person and Employee.

class Person has the << operator;

Now in class Employee I want this same behaviour plus some extra
behaviour for the fields which Employee has and Person doesn't;

I would have thought that since Person::operator() returns an ostream&
that assigning it to the existing o would be ok. However it isn't as
this compiler message informs me;

Now I understand this since it is not a member function as it is
declared as follows;

in person.hpp.

So how do I get round this problem? Basically I want to take the
returned ostream& from class Person's operator<< method and insert it
into the ostream& object in class Employee's operator<< method.

Thank,

John

On Mar 22, 8:54 am, "johnmmcparland" <johnmmcparl...@googlemail.com>
wrote:This operator is not a member. It's an overloaded operator<< that
takes a
Person& as its second argument. (Incidentally, the second parameter
should
probably be declared const.)

And here you're not *overriding* the base class' operator<< (you can't
override a non-member). You're defining another overloaded operator<<
that
takes an Employee& as its second argument.

In the first statement above you're trying to call the operator<< for
person as if it were a member function. It should be written

operator<<(o, static_cast<Person&>(emp));

or simply,

o << static_cast<Person&>(emp);

Incidentally, you don't need or want to assign the return value of
operator<<
back to o. It doesn't do what you think. operator<< returns a
reference to the
left hand operand so that you can chain the operator as you've done in
the
second statement above. This, the whole function body could be
written:

o << static_cast<Person&>(emp)
<< std::endl << "Started: " << emp.m_startDate
<< " Role: " << emp.m_job << "Dept: " << emp.m_department
<< std:: endl; << "Salary: " << emp.m_salary << std::endl;

Hope this helps.

--Nick

niklasb@microsoft.com wrote:
I agree with everything but the function body. You forgot to return the
ostream ;) like this:

return o
<< static_cast<Person&>(emp)
<< std::endl << "Started: " << emp.m_startDate
<< " Role: " << emp.m_job << "Dept: " << emp.m_department
<< std:: endl; << "Salary: " << emp.m_salary << std::endl;

But I'm sure that was just an oversight.

Cheers!


Adrian
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On 22 Mar, 16:59, Adrian Hawryluk <adrian.hawryluk-at-
gmail....@nospam.comwrote:Thanks Guys. The cast you suggested Nick was what I was looking for.
I had tried a different cast (Java style - (Person)emp) but this had
failed. Maybe its time I re-read my notes on C++ casts.

John

On Mar 22, 11:54 am, "johnmmcparland" <johnmmcparl...@googlemail.com>
wrote:

My way of dealing with this:

class Person
{
// :
virtual void Output(ostream& o) const
{
o << m_surname << ", " << m_firstname;
if (m_middlename != "")
o << m_middlename;
o << " (" << m_dob << ") " << "[";
// etc....
}
};

class Employee : public Person
{
// :
virtual void Output(ostream& o) const
{
Person::Output(o);
o << std::endl << "Started: " << m_startDate
<< " Role: " << m_job << "Dept: " << m_department
<< std:: endl << "Salary: " << m_salary << std::endl;
}
};


template<typename T>
ostream& operator<<(ostream& o, const T& t)
{
t.Output(o);
return o;
}

No ugly casting. Nothing needs to be a friend. Easily Extendable.

johnmmcparland wrote:Yet another alternative:

struct Streamable
{
virtual void output( std::ostream& ) const = 0;
};

std::ostream& operator<<( std::ostream& out,
const Streamable& s ) const
{
s.output( out );
return out;
}

And derive your Person from Streamable.

--
Ian Collins.

On Mar 22, 9:59 am, Adrian Hawryluk <adrian.hawryluk-at-
gmail....@nospam.comwrote:Yep. Thanks for catching it. :-)