I did not get right result for object f3 when I see the f3 using show_fraction() function in the below code: I am confused. I want to access the sum of f1 and f2 fraction that is stored in f3.But when i see the result in f3 it shows 0/0 in result using fraction_show() function

#include <iostream>
using namespace std;
class fraction
{
private:
    float numerator;
    float denominator;
    char ch;

public:
    void get_fraction()
    {
        cout << "\nEnter the fraction ( in format n/d ): ";
        cin >> numerator >> ch >> denominator;
    }

    void show_fraction()
    {
        cout << " Fraction is: " << numerator << "/" << denominator;
    }

    fraction sum(fraction &f, fraction &d) // returns sum of two fractions
    {
        fraction ff; // to hold the result
        // denominator of result
        ff.numerator = (f.numerator * d.denominator   f.denominator * d.numerator);
        ff.denominator = f.denominator * d.denominator; // numerator of result
        return ff;
    }
};

int main() // program to test the class
{
    fraction f1, f2, f3; // three objects of type fraction created
    char ch1;

    do
    {
        f1.get_fraction();
        f2.get_fraction();
        f3.sum(f1, f2);
        cout << "\nSum of ";
        f3.show_fraction();                            // shows the result ( wrong result)
        cout << "\nDo you want to continue ( y/n ): "; // If the user want to continue
        cin >> ch1;

    } while (ch1 != 'n');

    cout << endl;
    return 0;
} // end of main

CodePudding user response：

I don't like this style of coding, but at least this works

void sum(fraction &f, fraction &d) // calculates sum of two fractions
{
    numerator = (f.numerator * d.denominator   f.denominator * d.numerator);
    denominator = f.denominator * d.denominator;
}

then

f3.sum(f1, f2);

I would prefer something like this

friend fraction sum(fraction &f, fraction &d) // returns sum of two fractions
{
    fraction ff; // to hold the result
    // denominator of result
    ff.numerator = (f.numerator * d.denominator   f.denominator * d.numerator);
    ff.denominator = f.denominator * d.denominator; // numerator of result
    return ff;
}

with this version you need to change the way you call the function

f3 = sum(f1, f2);

I'm not sure which you were intending but you wrote something that was half way between the two versions.