mshiels@tmsoft.uucp (Michael A. Shiels) (08/16/89)
Article 17981 of comp.sys.ibm.pc:
Path: watmath!looking!brad
From: brad@looking.UUCP (Brad Templeton)
Newsgroups: comp.sys.ibm.pc
Subject: Re: How to tell the difference between 80386 virtual 8086 mode and 286 real mode?
Message-ID: <1884@looking.UUCP>
Date: 30 Jul 88 16:56:17 GMT
References: <1873@looking.UUCP> <1657@microsoft.UUCP>
Reply-To: brad@looking.UUCP (Brad Templeton)
Organization: Looking Glass Software Ltd.
Lines: 49
> use SMSW
This does sound like a good method. Another method, which I developed
on my own while waiting for replies, was to examine the length of the
instruction prefetch queue. This can even tell the difference between
an 8088 and and 8086 if you want it to. (8088 = 4, 8086,286 = 6, 386 = 8)
But that does sound like a rather scary way to do things, since Intel is
certainly free to change the length of this in future chips, or to put
in impure code checking into the prefetch hardware.
For those who want it,...
prefcount PROC NEAR
; must preserve ax. It does
mov bx,offset point
mov cx,0
mov dx,12 ; max iter
kloop:
nop
nop
; store "inc cx" at one of the nops
mov byte ptr cs:[bx],41h ; inc cx
point:
; execute a chain of nops, one that might be the "inc cx"
REPT 12
nop
ENDM
; restore the "inc cx" back to a nop
mov byte ptr cs:[bx],90h ; nop
; go on to the next byte
inc bx
; check we haven't gone too far
dec dx
jz pcdone
; keep looping until the inc was interpreted
jcxz kloop
pcdone:
; we already have the index of the first non-prefetched
; byte plus one (we incremented anyway). Now we use
; point-1 to add another to it because of the three byte
; instruction at the front
; the result is 6 for 286 and 8 for 386. (4 for 8088)
sub bx,offset point-1
ret
prefcount ENDP
--
Brad Templeton, Looking Glass Software Ltd. -- Waterloo, Ontario 519/884-7473