Re: Overload operator * as member function.

From:

Nick Hounsome <nick.hounsome@googlemail.com>

Newsgroups:

comp.lang.c++.moderated

Date:

Sun, 12 Jul 2009 05:56:22 CST

Message-ID:

<1af94c4a-60ef-4326-a896-4ab1d0748279@n30g2000vba.googlegroups.com>

On 11 July, 17:36, AY <techreposit...@gmail.com> wrote:

class A
{

                const A operator*(const A& rt, const A& lt)
                {
                        return A(rt.denominator()*lt.denominator(),
                                                        rt.numerator()*lt.numerator());
                }

As a member you only want the right hand parameter:

const A operator*(const A& rhs); // The lhs is implicitly "this".

Always implement binary operators as free functions as otherwise
expressions like ( 3 + A(1,2) ) wont compile because the compiler will
not try to convert the 3 to an A when looking for a function match.

Always implement binary operators (op) in terms of a member method
operator op= as follows:

A& A::operator *= (const A& rhs) { blah blah; return *this; }
A operator* (const A& lhs, const A& rhs) { A temp(lhs); temp *= rhs;
return temp; }

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