New to board problems with overloading

[Lurker]

You're close with your '+' function, but you're adding the fraction to itself. I think:

value.numerator = (numerator * rat.denominator) + (rat.numerator * denominator);
value.denominator = denominator * rat.denominator;

will work. The left side of the + sign should be able to be represented by the 'private' variables of your class. I did the rational class about two months ago, and the good news is once you get one to work you'll be able to do the rest in no time.

[Turbo02]

Im starting to understand this now somewhat. My question is

value.numerator = (numerator * rat.denominator) + (rat.numerator * denominator);
value.denominator = denominator * rat.denominator;

how do you know where to place the rat.numerator or rat.denominator

[golfinguy4]

Originally posted by Stoned_Coder
No. operator + should always return an object and not a reference. Remember this rule. It will serve you well.

Yep, you are right. I was thinking of the = operator.

[Lurker]

The rat.numerator and rat.denominator are on the right side of the operator. So, if you are adding:

solution = 2/3 + 3/4

rat.numerator is 3, and rat.denomator is 4.

[elad]

or you could try this:

class Rational
{
   public:
     //do dah
     friend Rational & operator * (const Rational & lhs, const Rational & rhs);
     //more
};

Rational  operator*(const Rational & lhs, const Rational & rhs)
{
   Rational result;
   result.numerator = lhs. numerator * rhs.numerator;
   result.denominator = lhs.denominator * rhs.denominator;
   return result;
}

you can do a similar syntax (specify the parameters on the left and right hand side) for +. Of course, you need to determine the lowest common denominator just like you do when doing this by hand before you can add the numerators and denominators.

[Turbo02]

Originally posted by Lurker
The rat.numerator and rat.denominator are on the right side of the operator. So, if you are adding:

solution = 2/3 + 3/4

rat.numerator is 3, and rat.denomator is 4.

Thanx for your help I appreciate it. I now understand. Im going to take my final in an hour everyone wish me luck hahahah

[Stoned_Coder]

Most modern compilers support something known as the return value optimisation.
this

Rational  operator*(const Rational & lhs, const Rational & rhs)
{
   Rational result;
   result.numerator = lhs. numerator * rhs.numerator;
   result.denominator = lhs.denominator * rhs.denominator;
   return result;
}

is better written as:-

Rational operator*(const Rational& lhs,const rational& rhs)
{
    return Rational ((lhs.numerator*rhs.numerator),(lhs.denominator*rhs.denominator));
}

This eliminates the otherwise needed temporary.