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