Fizzbuzz in C accessing disk

Question

I have made a simple fizzbuzz program in C, just for practicing some aspects of the C language. I am still learning it, I am coming from higher level languages.

While running the program itself, with higher numbers, I noticed my HDD LED blinking on my PC. Did I do something wrong in the code, or is it writing to swap, or both?

(Running on Pop._OS 20,04. compiled with gcc 9.3,0, high number as in 30 million)

Thanks in advance!

Code:

#include <stdio.h>
#include <string.h>
#include <stdlib.h>

void fizzbuzz(int max){
    for(int i = 0; i <= max; ++i){
        if(i%3==0)
            printf("Fizz\n");
            
        if(i%5==0)
            printf("Buzz\n");
            
        if(i%15==0)
            printf("FizzBuzz\n");
            
        else
            printf("%d\n", i);
    }
}

int main( int argc, char *argv[] ){
    // argc: Count of arguments passed
    // argv: Pointer to arguments
    
    // Checking if arguments were passed, and checking the arguments themselves
    
    
    if(argc>1){
        if(strncmp(argv[1], "-max", 4) == 0){
            fizzbuzz(atoi(argv[2]));
        }
        else{
            printf("\nUsage: fizzbuzz_adv -max <limit of fizzbuzz>");
        }
    }
    else{
        printf("\nUsage: fizzbuzz_adv -max <limit of fizzbuzz>");
    }
    
    return 0;
}

Answer 1

From the idea of MM and Barmar I installed an I/O tracking utility (iotop) and it seems like gnome-terminal is writing to disk, instead of the program itself.