minimal init daemon by rich felker, author of musl libc

init.c

#define _XOPEN_SOURCE 700
#include <signal.h>
#include <unistd.h>

int main()
{
	sigset_t set;
	int status;

	if (getpid() != 1) return 1;

	sigfillset(&set);
	sigprocmask(SIG_BLOCK, &set, 0);

	if (fork()) for (;;) wait(&status);

	sigprocmask(SIG_UNBLOCK, &set, 0);

	setsid();
	setpgid(0, 0);
	return execve("/etc/rc", (char *[]){ "rc", 0 }, (char *[]){ 0 });
}

For those confused, all this code does is:

1. Get the process ID of the current process.
2. If the process ID is not 1, return 1 (error).
3. Fill the signal set with all current signals.
4. Block all signals via the SIG_BLOCK constant.
5. If a fork is successful, wait for the child to exit.
6. Unblock all signals.
7. Set the session ID of the process to the process ID.
8. Execute the /etc/rc script.

Is it not possible to straight up execute the /etc/rc script? Just by adding init=/etc/rc to the kernel command line?

Just go outside instead of turning on your computer. It uses 0B of disk space

I'm 100% sure this is way too slow. We need to write this in assembly. So that it'll fit in 1 sector of a drive.

I like shell scripts that don't need to be recompiled... A mount -a && /etc/rc works just fine

Is it not possible to straight up execute the /etc/rc script? Just by adding init=/etc/rc to the kernel command line?

You need a always running daemon, who will be reap a zombie. It is single work for init.
If you redirect this job to /etc/rc, you got a systemd again.
You cannot start an another daemon from /etc/rc for this. Zombie go to process with PID 1. /etc/rc have a PID 1 in your case.