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

In this example, we will learn to Add Two Distances (in inch-feet) Using Structures.

To understand this example, you should have the knowledge of the following C++ programming topics:

## Example: Program to Add Distances Using Structures

```
#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)
{
++ sum.feet;
sum.inch -= 12;
}
cout << endl << "Sum of distances = " << sum.feet << " feet " << sum.inch << " inches";
return 0;
}
```

**Output**

Enter 1st distance, Enter feet: 6 Enter inch: 3.4 Enter information for 2nd distance Enter feet: 5 Enter inch: 10.2 Sum of distances = 12 feet 1.6 inches

## working

In this program, a structure `Distance`

containing two data members
(`inch` and `feet`) is declared to store the distance in inch-feet
system.

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

The `if..else`

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

## Next Example

We hope that this Example helped you develop better understanding of the concept of "Add Two Distances (in inch-feet system) using Structures" in C++.

**Keep Learning : )**

In the next Example, we will learn about C++
`Add Two Complex Numbers by Passing Structure to a Function`

.