Count even/odd numbers using pointer Dry Run in C

Count even/odd numbers using pointer is an interactive C dry run visualizer from the Pointersarrays programs section. Study the source code, then use the execution controls to follow each step, variable update, highlighted line, and console output.

This page provides a browser-based dry run with source-code highlighting, auto-scroll, voice narration controls, and execution output for learning the program step by step.

Count even/odd numbers using pointer Program Code

#include <stdio.h>
void countEvenOdd(int *ptr, int n, int *even, int *odd) {
    int i;
    *even = 0;
    *odd = 0;
    
    for(i = 0; i < n; i++) {
        if(*(ptr + i) % 2 == 0) {
            (*even)++;
        } else {
            (*odd)++;
        }
    }
}

int main() {
    int i;
    int arr[] = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12};
    int n = sizeof(arr)/sizeof(arr[0]);
    int evenCount, oddCount;
    
    countEvenOdd(arr, n, &evenCount, &oddCount);
    
    printf("Array: ");
    for(i = 0; i < n; i++) {
        printf("%d ", arr[i]);
    }
    printf("\nEven numbers: %d", evenCount);
    printf("\nOdd numbers: %d", oddCount);
    
    return 0;
}

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