C++ Program to Add Two Distances (in inch-feet) System Using Structures

To add two distances in the inch-feet system using structures, we can create a structure to represent a distance with two members: feet and inch.

Here's the C++ program :

Add Distances Using Structures :

Program :

#include <iostream>
using namespace std;

struct Distance {
    int feet;
    float inch;
}d1 , d2, sum;

int main() {
    cout << "Enter 1st distance," << endl;
    cout << "Enter feet: ";
    cin >> d1.feet;
    cout << "Enter inch: ";
    cin >> d1.inch;

    cout << "\nEnter information for 2nd distance" << endl;
    cout << "Enter feet: ";
    cin >> d2.feet;
    cout << "Enter inch: ";
    cin >> d2.inch;

    sum.feet = d1.feet+d2.feet;
    sum.inch = d1.inch+d2.inch;

    // changing to feet if inch is greater than 12
    if(sum.inch > 12) {
        // extra feet
        int extra = sum.inch / 12;

        sum.feet += extra;
        sum.inch -= (extra * 12);
    } 

    cout << endl << "Sum of distances = " << sum.feet << " feet  " << sum.inch << " inches";
    return 0;
}

Output :

Enter 1st distance,
Enter feet: 12
Enter inch: 6
Enter information for 2nd distance
Enter feet: 6
Enter inch: 2
Sum of distances = 18 feet  8 inches

In this program, a structure Distance containing two data members (inch and feet) is declared to store the distance in the inch-feet system.

Here, two structure variables d1 and d2 are created to store the distance entered by the user. And, the sum variable stores the sum of the distances.

The if statement is used to convert inches to feet if the value of inch of sum variable is greater than 12:

The int variable extra stores the extra feet gained due to the value of inch being greater than 12. This is obtained from the quotient of division between sum.inch and 12.

We then add the extra feet to sum.feet.

The true value of sum.inch is then calculated by subtracting extra * 12 from its initial value.

Another Example :

Program :

#include <iostream>

using namespace std;

struct Distance
{
    int feet;
    int inch;
};

void displayDistance(Distance d);

Distance addDistance(Distance d1, Distance d2);

int main()
{
    Distance d1 = {10, 6};
    Distance d2 = {5, 10};

    Distance sum = addDistance(d1, d2);

    cout << "First distance: ";
    displayDistance(d1);
    cout << "Second distance: ";
    displayDistance(d2);
    cout << "Sum of distances: ";
    displayDistance(sum);

    return 0;
}

void displayDistance(Distance d)
{
    cout << d.feet << " feet " << d.inch << " inches" << endl;
}

Distance addDistance(Distance d1, Distance d2)
{
    Distance sum;
    sum.feet = d1.feet + d2.feet;
    sum.inch = d1.inch + d2.inch;

    if (sum.inch >= 12)
    {
        sum.feet++;
        sum.inch -= 12;
    }

    return sum;
}

Output :

First distance: 10 feet 6 inches
Second distance: 5 feet 10 inches
Sum of distances: 16 feet 4 inches

In this program, we define a structure Distance with two members feet and inch. The function displayDistance() is used to display a distance in the inch-feet system. The function addDistance() takes two distances as arguments and returns their sum.

In the main() function, we create two distances d1 and d2 and initialize them with values. We then call the addDistance() function to add these distances and store the result in the sum variable. Finally, we call the displayDistance() function to display all three distances.

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