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C Program to Add N Distances Given in inch-feet System using Structures

Last Updated : 23 Jul, 2025
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Given an array arr[] containing N distances of inch-feet system, such that each element of the array represents a distance in the form of {inch, feet}. The task is to add all the N inch-feet distances using structures.

Examples:

Input: arr[] = { { 10, 3.7 }, { 10, 5.5 }, { 6, 8.0 } }; 
Output: Feet Sum: 27 Inch Sum: 5.20
 
Input: arr[] = { { 1, 1.7 }, { 1, 1.5 }, { 6, 8 } }; 
Output: Feet Sum: 8 Inch Sum: 11.20

Approach

1. Traverse the struct array arr and find the summation of all the inches of the given set of N distances as:

feet_sum = feet_sum + arr[i].feet;
inch_sum = inch_sum + arr[i].inch;

2. If the sum of all the inches (say inch_sum) is greater than 12, then convert the inch_sum into feet because

1 feet = 12 inches

3. Therefore update inch_sum to inch_sum % 12. Then find the summation of all the feet (say feet_sum) of N distances and add inches_sum/12 to this sum.

4. Print the feet_sum and inch_sum individually.

feet_sum = feet_sum + arr[i].feet;
inch_sum = inch_sum + arr[i].inch;

Program to add two distances in the inch-feet System

Below is the implementation of the above approach: 

C 
// C program for the above approach
#include "stdio.h"

// Struct defined for the inch-feet system
struct InchFeet {
    // Variable to store the inch-feet
    int feet;
    float inch;
};

// Function to find the sum of all N
// set of Inch Feet distances
void findSum(struct InchFeet arr[], int N)
{
    // Variable to store sum
    int feet_sum = 0;
    float inch_sum = 0.0;

    int x;

    // Traverse the InchFeet array
    for (int i = 0; i < N; i++) {
        // Find the total sum of
        // feet and inch
        feet_sum += arr[i].feet;
        inch_sum += arr[i].inch;
    }

    // If inch sum is greater than 11
    // convert it into feet
    // as 1 feet = 12 inch
    if (inch_sum >= 12) {
        // Find integral part of inch_sum
        x = (int)inch_sum;

        // Delete the integral part x
        inch_sum -= x;

        // Add x%12 to inch_sum
        inch_sum += x % 12;

        // Add x/12 to feet_sum
        feet_sum += x / 12;
    }

    // Print the corresponding sum of
    // feet_sum and inch_sum
    printf("Feet Sum: %d", feet_sum);
    printf("Inch Sum: %.2f", inch_sum);
}

// Driver Code
int main()
{
    // Given set of inch-feet
    struct InchFeet arr[]
        = { { 10, 3.7 }, { 10, 5.5 }, { 6, 8.0 } };

    int N = sizeof(arr) / sizeof(arr[0]);

    // Function Call
    findSum(arr, N);

    return 0;
}
Output:
Feet Sum: 27
Inch Sum: 5.20

The complexity of the above method

Time Complexity: O(N), where N is the number of inch-feet distances.


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kartik
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