Write the function in C ++ to reduce the numerator and denominator in the Rat cl

Question

Write the function  in C ++ to reduce the numerator and denominator in the Rat class to lowest terms.

// class for rational numbers

#include <iostream>

using namespace std;

class Rat{

private:

int n;

int d;

public:

// constructors

// default constructor

Rat(){

n=0;

d=1;

}

// 2 parameter constructor

Rat(int i, int j){

n=i;

d=j;

}

// conversion constructor

Rat(int i){

n=i;

d=1;

}

//accessor functions (usually called get() and set(...) )

int getN(){ return n;}

int getD(){ return d;}

void setN(int i){ n=i;}

void setD(int i){ d=i;}

//arithmetic operators

Rat operator+(Rat r){

Rat t;

t.n=n*r.d+d*r.n;

t.d=d*r.d;

return t;

}

2

// 2 overloaded i/o operators

friend ostream& operator<<(ostream& os, Rat r);

friend istream& operator>>(istream& is, Rat& r);

}; //end Rat

// operator<<() is NOT a member function but since it was declared a friend of Rat

// it has access to its private parts.

ostream& operator<<(ostream& os, Rat r){

os<<r.n<<" / "<<r.d<<endl;

return os;}

// operator>>() is NOT a member function but since it was declared a friend of Rat

// it has access to its provate parts.

istream& operator>>(istream& is, Rat& r){

is>>r.n>>r.d;

return is;

}

int main(){

Rat x(1,2), y(2,3), z;

z=x+y;

cout<<z;

x.setN(3);

y.setD(2);

z=x+y;

cout<<z;

cin>>x;

cout<<x;

z= x+5;

cout<<z;

system("pause");

return 0;

}

Explanation / Answer

void reduce()
{
// first find which one is bigger out of n and d.
int max = (n>d?n:d);
int gcd = 1;
// intialize gcd with 1.
for(int i=max; i>=1; i--)
{
// start checking for big factor that commonly divides n and d;
// once it found break it.
if(n%i==0 && d%i==0) { gcd = i; break; }
}
// dividing both numerator and denominator with gcd.
n = n/gcd;
d = d/gcd;
}

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