Solved

# class rational

Posted on 2004-08-07
563 Views
Can you comment on the effectiveness of this code?  I tried to create a class for rational numbers with overloaded operators which apply to my rational type class?  This is a homework assignment, but I wanted to make sure it looked ok.

class Rational {
public:
Rational(int t = 0, int n = 1);

friend Rational operator+ (const Rational& lVal, const Rational& rVal);
friend Rational operator- (const Rational& lVal, const Rational& rVal);
friend Rational operator* (const Rational& lVal, const Rational& rVal);
friend Rational operator/ (const Rational& lVal, const Rational& rVal);

Rational& operator+= (const Rational& rVal);
Rational& operator-= (const Rational& rVal);
Rational& operator*= (const Rational& rVal);
Rational& operator/= (const Rational& rVal);

friend int operator<  (const Rational& lVal, const Rational& rVal);
friend int operator<= (const Rational& lVal, const Rational& rVal);
friend int operator== (const Rational& lVal, const Rational& rVal);
friend int operator!= (const Rational& lVal, const Rational& rVal);
friend int operator>= (const Rational& lVal, const Rational& rVal);
friend int operator>  (const Rational& lVal, const Rational& rVal);

friend ostream& operator<< (ostream& os, const Rational& rat);
friend istream& operator>> (istream& is, Rational& rat);

private:
int teller, number;

void reduce();
int  number(int, int);
};

//Constructors

Rational::Rational(int t, int n): teller(t), number(n) {
reduce();
}

Rational Rational::operator -() const {
return Rational(-teller, number);
}

Rational operator+ (const Rational& lVal, const Rational& rVal) {
Rational result;

result.teller = lVal.teller * rVal.number + lVal.number * rVal.teller;
result.number = lVal.number * rVal.number;

result.reduce();
return result;
}

Rational operator- (const Rational& lVal, const Rational& rVal) {
Rational result;
result = lVal + -rVal;

result.reduce();
return result;
}

Rational operator* (const Rational& lVal, const Rational& rVal) {
Rational result;
result.teller = lVal.teller * rVal.teller;
result.number = lVal.number * rVal.number;

result.reduce();
return result;
}

Rational operator/ (const Rational& lVal, const Rational& rVal) {
Rational result;
result.teller = lVal.teller * rVal.number;
result.number = lVal.number * rVal.teller;
result.reduce();
return result;
}

Rational& Rational::operator+= (const Rational& rVal) {
teller = teller * rVal.number + number * rVal.teller;
number = number * rVal.number;

reduce();
return *this;
}

Rational& Rational::operator-= (const Rational& rVal) {
this->operator+=(-rVal);
reduce();
return *this;
}

Rational& Rational::operator*= (const Rational& rVal) {
teller = teller * rVal.teller;
number = number * rVal.number;

reduce();
return *this;
}

Rational& Rational::operator/= (const Rational& rVal) {
teller = teller * rVal.number;
number = number * rVal.teller;

reduce();
return *this;
}

int operator<  (const Rational& lVal, const Rational& rVal) {
return(lVal.toDouble() < rVal.toDouble());
}

int operator<= (const Rational& lVal, const Rational& rVal) {
return(lVal < rVal || lVal == rVal);
}

int operator== (const Rational& lVal, const Rational& rVal) {
return lVal.teller * rVal.number == lVal.number * rVal.teller;
}

int operator!= (const Rational& lVal, const Rational& rVal) {
return !(lVal == rVal);
}

int operator>= (const Rational& lVal, const Rational& rVal) {
return(lVal > rVal || lVal == rVal);
}

int operator>  (const Rational& lVal, const Rational& rVal) {
return(lVal.toDouble() > rVal.toDouble());
}

istream& operator>> (istream& is, Rational& rat) {
int t, n;
char deel;
is >> t >> deel >> n;
rat = Rational(t,n);
return is;
}

int main() {
clrscr();
Rational b1;
Rational b2;

cout << "Input the number in the form (a/b): ";
cin >> b1 >> b2;

// Test for +, -, *, /
cout << b1 << " + " << b2 << " = " << b1 + b2 << endl;
cout << b1 << " - " << b2 << " = " << b1 - b2 << endl;
cout << b1 << " * " << b2 << " = " << b1 * b2 << endl;
cout << b1 << " / " << b2 << " = " << b1 / b2 << endl;

// test program for overloaded operators
cout << b1 << " <  " << b2;
(b1 <  b2) ? cout << " [X]" << endl : cout << " [ ]" << endl;

cout << b1 << " <= " << b2;
(b1 <= b2) ? cout << " [X]" << endl << endl : cout << " [ ]" << endl << endl;

cout << b1 << " == " << b2;
(b1 == b2) ? cout << " [X]" << endl : cout << " [ ]" << endl;

cout << b1 << " != " << b2;
(b1 != b2) ? cout << " [X]" << endl << endl : cout << " [ ]" << endl << endl;

cout << b1 << " >= " << b2;
(b1 >= b2) ? cout << " [X]" << endl : cout << " [ ]" << endl;

cout << b1 << " >  " << b2;
(b1 >  b2) ? cout << " [X]" << endl : cout << " [ ]" << endl << endl;

getch();
return 0;
}
0
Question by:rumi54
[X]
###### Welcome to Experts Exchange

Add your voice to the tech community where 5M+ people just like you are talking about what matters.

• Help others & share knowledge
• Earn cash & points
• Learn & ask questions
• 2
• 2

LVL 55

Expert Comment

ID: 11743145
Everything seems to be ok, although I don't find Rational::reduce() implementation.
0

Author Comment

ID: 11743155
The GCF function...
0

LVL 55

Accepted Solution

Jaime Olivares earned 500 total points
ID: 11743264
LOL. Yes I know what means but maybe you can optimize something there.
Your comparison (<, >, ==, etc.) operators could be "inline" to enhance performance.
Your division operators have not "divide by zero" exception management.
0

Author Comment

ID: 11743283
I figured you know what it means.  :)  I thought I left it out completely.  Thanks.
0

## Featured Post

Question has a verified solution.

If you are experiencing a similar issue, please ask a related question

In days of old, returning something by value from a function in C++ was necessarily avoided because it would, invariably, involve one or even two copies of the object being created and potentially costly calls to a copy-constructor and destructor. A…
This article will show you some of the more useful Standard Template Library (STL) algorithms through the use of working examples.  You will learn about how these algorithms fit into the STL architecture, how they work with STL containers, and why t…
The viewer will learn how to pass data into a function in C++. This is one step further in using functions. Instead of only printing text onto the console, the function will be able to perform calculations with argumentents given by the user.
The viewer will learn how to use the return statement in functions in C++. The video will also teach the user how to pass data to a function and have the function return data back for further processing.
###### Suggested Courses
Course of the Month4 days, 9 hours left to enroll

#### 689 members asked questions and received personalized solutions in the past 7 days.

Join the community of 500,000 technology professionals and ask your questions.