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From: dhinds@portia.Stanford.EDU (David Hinds)
Newsgroups: comp.os.msdos.programmer
Subject: Re: Detecting an 80486
Message-ID: <1990Jul22.185546.136@portia.Stanford.EDU>
Date: 22 Jul 90 18:55:46 GMT
References: <26a858b9@ralf>
Organization: AIR, Stanford University
Lines: 23

In article <26a858b9@ralf> Ralf.Brown@B.GP.CS.CMU.EDU writes:
>In article <1990Jul20.230335.22816@ddsw1.MCS.COM>, andyross@ddsw1.MCS.COM (Andrew Rossmann) wrote:
>}  For the 286, it tries the SMSW DX instruction (whatever that does.)
>}  The code come from a program by Robert Collins, and I just made a few
>}changes to fit INFOPLUS (and fix the 386 problem!)
>
>Unfortunately, your program will fail miserably on my system (and about a
>million others), because I am running QEMM-386.  QEMM runs in protected mode
>with DOS in virtual-86 mode.  Any invalid opcode exception that your
>program generated invokes the QEMM exception handler, NOT YOURS.  And the
>QEMM exception handler gives the choice of terminating or rebooting.  I
>expect similar behavior from 386^Max, CEMM, AllCharge386, MICEMM, ....

   The SMSW instruction will fail if executed on less than a 286, so it
is a safe test even in the presence of QEMM.  As a bonus, the low-order bit
of the MSW (machine status word) is the protected-mode flag.  The only way
it can be set under DOS is if a 386 is running in virtual-86 mode.  So, if
the bit is set, you are on a 386.  If it isn't, you are either on a 286 or
386 in real mode, and it is safe to try a 386 instruction and trap the
interrupt.

 -David Hinds
  dhinds@popserver.stanford.edu