2010-1-24 Linux Local Descriptor Table

Is 32bit Intel ELF 0x80**** adreeses is default? nope. You can setup
your own. Compiler will not see thembut you can do it.
Setup LDT and you will see it.

use32
mov dword [0] ,"Hall"
mov dword [4] ,"Ball"
mov dword [8] ,"Mall"
mov dword [12],0x00000000

yes everything starts from 0x0

#include <stdlib.h>

#include <stdio.h>
#include <sys/syscall.h>
#include <sys/types.h>
#include <asm/ldt.h>

char new_segment[16];

int main()
{
	int r;
	
	struct user_desc *ldt;
	
	ldt = (struct user_desc*)malloc(sizeof(struct user_desc));
	
	ldt->entry_number = 0;
	ldt->base_addr = ((unsigned long)&new_segment);
	ldt->limit = 16;
	ldt->seg_32bit = 0x1;
	ldt->contents = 0x0;
	ldt->read_exec_only = 0x0;
	ldt->limit_in_pages = 0x0;
	ldt->seg_not_present = 0x0;
	ldt->useable = 0x1;
	
	printf("Start\n");
	r = syscall( __NR_modify_ldt, 1 , ldt , sizeof(struct user_desc) );
	if ( r == -1 )
	{
		printf("Sorry\n");
		exit( 0 );
	}
	asm("pushl %ds");
	asm("movl $0x7, %eax"); /* 0111: 0-Index 1-Using the LDT table 11-RPL of 3 */
	asm("movl %eax, %ds");	
	asm(".byte 0xc7,0x5,0x0,0x0,0x0,0x0,0x48,0x61,
	0x6c,0x6c,0xc7,0x5,0x4,0x0,0x0,0x0,
	0x42,0x61,0x6c,0x6c,0xc7,0x5,0x8,0x0,
	0x0,0x0,0x4d,0x61,0x6c,0x6c,0xc7,0x5,
	0xc,0x0,0x0,0x0,0x0,0x0,0x0,0x0");
	asm("popl %ds");
	printf("End\n");
	
	printf("Segment [%s]\n",new_segment);
	
	free( ldt );
	
	return 0;
}

asm(".byte ... ") is code.bin

Compile:
fasm code.asm code.bin

gcc main.c -o main