C - getopt switch while loop and default case for invalid options

Question

I want to get the program to hit the default case if an invalid option/command is typed, but it doesn't even enter the while loop. I'm wondering what I'm doing wrong, and what I would need to change to get it working, it only works if a correct case is used. :)

#include <stdio.h>
#include <stdlib.h>
#include <getopt.h>
#include <string.h>
int main (int argc, char *argv[]){

const char *optstring = "rs:pih";
char option;

while ((option = getopt(argc, argv, optstring)) != EOF) {

printf("I'm in the while loop!");

    switch (option) {
        case 'r':
            break;
        case 's':
            printf("Second Argument: %s", optarg);
            break;
        case 'p':
            printf("p");
            break;
        case 'i':
            printf("i");
            break;
        case 'h':
            printf("h");
            break;
        default:
            printf("nothing");

    }

 }

 return 0;
}

Answer 1

From the comments:

./program_name d doesn't have a flag. Try: ./program_name -r -s foo -p

You could also print the remaining options that weren't parsed via:

for (int i = optind; i < argc; i++) {
    printf("remaining argument[%d]: %s\n", i, argv[i]);
}