C Program to Find Sum of Even and Odd Numbers in an Array

Given a list of integers, separate them into two groups — even numbers (divisible by 2) and odd numbers (not divisible by 2) — and report the sum and count of each group. This is a classic array-scanning problem that demonstrates the modulo operator, accumulator variables, and loop-based input. The original post had no code; this rewrite provides a complete, working C89-compliant implementation.

How It Works

Read N integers into an array.
For each element, test a[i] % 2 == 0.
Add to even_sum (and increment even_count) or to odd_sum (and increment odd_count).
Print both sums and counts.

i	a[i]	a[i] % 2	Group	even_sum	odd_sum
0	1	1	Odd	0	1
1	2	0	Even	2	1
2	3	1	Odd	2	4
3	4	0	Even	6	4
4	5	1	Odd	6	9

Result for [1,2,3,4,5]: even_sum=6, odd_sum=9 ✓

C Program: Sum of Even and Odd Numbers

/* Read N integers and print the sum of even and odd numbers separately
 * Compile: gcc -ansi -Wall -Wextra sumeodd.c -o sumeodd */
#include <stdio.h>
#define MAX 100

int main(void)
{
    int a[MAX], n, i;
    int even_sum = 0, odd_sum = 0;
    int even_count = 0, odd_count = 0;

    printf("How many integers? ");
    if (scanf("%d", &n) != 1 || n < 1 || n > MAX) {
        printf("Enter a number between 1 and %d.\n", MAX);
        return 1;
    }

    printf("Enter %d integers:\n", n);
    for (i = 0; i < n; i++) {
        if (scanf("%d", &a[i]) != 1) {
            printf("Invalid input.\n");
            return 1;
        }
    }

    for (i = 0; i < n; i++) {
        if (a[i] % 2 == 0) {
            even_sum += a[i];
            even_count++;
        } else {
            odd_sum += a[i];
            odd_count++;
        }
    }

    printf("\nEven numbers (%d): sum = %d\n", even_count, even_sum);
    printf("Odd  numbers (%d): sum = %d\n", odd_count,  odd_sum);

    return 0;
}

How to Compile and Run

gcc -ansi -Wall -Wextra sumeodd.c -o sumeodd
./sumeodd

Sample Output

How many integers? 10
Enter 10 integers:
1 2 3 4 5 6 7 8 9 10

Even numbers (5): sum = 30
Odd  numbers (5): sum = 25

How many integers? 5
Enter 5 integers:
3 7 2 9 4

Even numbers (2): sum = 6
Odd  numbers (3): sum = 19

Code Explanation

Two accumulators, two counters — even_sum/odd_sum accumulate the running totals; even_count/odd_count track how many of each kind there are. Initializing both to 0 at declaration is important — uninitialized local variables hold garbage in C.
Single-pass loop — the array is read in one loop and classified in a second loop. These could be merged into one loop (reading and classifying in the same iteration), but splitting them makes the logic clearer. For this problem, either approach is correct.
Bounds validation — checking n < 1 || n > MAX after reading prevents array overflow. Without this, entering n=200 with a MAX=100 array causes undefined behavior (writing past the array boundary).
#define MAX 100 — using a named constant instead of a magic number means changing the array size requires editing exactly one line. Always prefer named constants over magic numbers in production code.

What This Program Teaches

Accumulator pattern — initializing a variable to zero before a loop and adding to it inside the loop is the canonical way to compute sums and counts in C. It appears in averaging, totalling invoices, computing checksums, and signal integration.
Modulo for classification — % 2 partitions integers into exactly two groups. The same technique partitions into N groups with % N. Classic uses: round-robin scheduling, hash-bucket assignment, striped array access.
Input validation before use — check both the return value of scanf and the logical validity of the read value (range check). Reading values first, validating second, is the standard idiom.

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C Program to Find the Sum of Even and Odd Numbers