linux-kernel - Re: [i386] Questions regarding provisional page tables initialization

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Date:	Mon, 2 Jul 2007 13:23:03 +0300
From:	"Ahmed S. Darwish" <darwish.07@...il.com>
To:	Andreas Schwab <schwab@...e.de>
Cc:	Jeremy Fitzhardinge <jeremy@...p.org>, linux-kernel@...r.kernel.org
Subject: Re: [i386] Questions regarding provisional page tables initialization

Hi Andreas,

On Mon, Jul 02, 2007 at 11:18:08AM +0200, Andreas Schwab wrote:
> "Ahmed S. Darwish" <darwish.07@...il.com> writes:
> 
> > yes, but isn't the displacement here (0x007) a _bytes_ displacement ?. so
> > effectively, %ecx now contains physical address of pg0 + 7bytes. Is it A 
> > meaningful place/address ?.
> 
> It's not pg0 + 7bytes, it is pg0 plus 3 flag bits.  Since a page address
> is always page aligned, the low bits are reused for flags.
> 

I'm sure there's a problem in _my_ understanding, but isn't the displacement
- as specified by AT&T syntax - represented in bytes ?. I've wrote a small
assembly function to be sure:

.data
integer:
	.string "%d\n"

.text
test_func:
	push	%ebp
	mov	%esp, %ebp
	push	0x008(%ebp)   ## 8 bytes displacement (the first arg), right ?
	push	$integer
	call 	printf
	mov	%ebp, %esp
	pop	%ebp
	ret

The above method works fine and prints "5" to stdout by the code:

.global	main
main:
	mov	$5, %eax
	push	%eax
	call	test_func

	movl	$1, %eax
	movl	$0, %ebx
	int	$0x80

now back to head.S code:
 	leal    0x007(%edi),%ecx	/* Create PDE entry */

Isn't the above line the same condition (bytes, not bits displacement) ?. 
Thanks for your patience !.

(For other kind replies, don't understand me wrong. I did my homework and
 studied the pte format before asking ;). It's just the bytes/bits issue 
 above that confuses me).

-- 
Ahmed S. Darwish
HomePage: http://darwish.07.googlepages.com
Blog: http://darwish-07.blogspot.com
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