Re: [9fans] calculating the gdt pointer in 32 bit (protected) mode

[andrey mirtchovski <andrey@lanl.gov>]

the instruction is:

	BYTE		$0x0f
	BYTE 	$0x01
	BYTE 	$0x16
	LONG	$tgdptr-KZERO(SB)

when traced it resembles:

	MOVL	([ESI]+($tgdptr-KZERO(SB))), GDTR

(the 0x16 tells it that the address if offset from ESI)

if instead one assembles:

	MOVL	$tgdptr-KZERO(SB), GDTR

(as it is written in the comments above the original
instruction) then the third byte becomes 0x15 and
it starts thinking that the pointer is actually a 32 bit
displacement, which is even less correct (i think)..

i'm going to start debugging using the proven trial-
error method and will see which takes me to the
correct GDT pointer..

am i correct in my assumption that i am unable to
modify the ESI register? otherwise i'd be tempted
to put the pointer value in ESI and execute the
instruction with 0 displacement from ESI..

i could of course be terribly wrong in everything i've
done so far...

andrey