How to write an operating system

This is a quick guide to creating a ‘Hello World’ Operating system that can be booted by Grub First of all you will need to create the assembly file the contains the entry point for grub. Call this ‘start.S‘

/*
* initial stack
*/
.data

.globl	_os_stack
.align 	4, 0x90
.space 	0x100
_os_stack:
.long 0

/*
* lets roll
*/
.text

.globl 	_start
.align 	4, 0x90
_start:
jmp	boot_entry

/*
* multiboot compliant header
* the layout of this thing depends on image format
*/

.align 4, 0x90
boot_hdr:

.long	0x1BADB002		/* magic */
.long   0x00000000		/* no flags if ELF */
.long	0-0x1BADB002-0x00000000	/* checksum */

/* the actual code starts here ... */
boot_entry:
/*
* clear NT, learned the hard way...
*/
pushl 	$0
popfl

/*
* setup our stack
*/
lea   	_os_stack, %edx
movl  	%edx, %esp

/*
* mulitboot compliant loader (read: grub)
* should put magic value in %eax
*/
pushl	%eax

/*
* clear bss
*/
xorl  	%eax, %eax
movl  	$edata, %edi
movl  	$end, %ecx
subl  	%edi, %ecx
cld
rep
stosb

/*
* call our C code initialization ...
*/
pushl 	%ebx
call  	main

/*
* NOTREACHED
*/
darn:
incw	(0xb8000)
jmp	darn

Now you will need to create your ‘main.c‘ program

int main(){
	char * vidmem = (char*)0xB8000;
	vidmem[0]  = 'H';
	vidmem[1]  = 0x7;
	vidmem[2]  = 'e';
	vidmem[3]  = 0x7;
	vidmem[4]  = 'l';
	vidmem[5]  = 0x7;
	vidmem[6]  = 'l';
	vidmem[7]  = 0x7;
	vidmem[8]  = 'o';
	vidmem[9]  = 0x7;
	vidmem[10] = ' ';
	vidmem[11] = 0x7;
	vidmem[12] = 'W';
	vidmem[13] = 0x7;
	vidmem[14] = 'o';
	vidmem[15] = 0x7;
	vidmem[16] = 'r';
	vidmem[17] = 0x7;
	vidmem[18] = 'l';
	vidmem[19] = 0x7;
	vidmem[20] = 'd';
	vidmem[21] = 0x7;
	vidmem[22] = '!';
	vidmem[23] = 0x7;
}

Now we need to compile our program

gcc -nostdinc -c main.c -o main.o
gcc -nostdinc -c start.S -o start.o
ld -nostdinc -Ttext 0x100000 main.o start.o -o kernel

…And thats it, you add an entry to your grub ‘menu.lst‘