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From: kaufman@neon.Stanford.EDU (Marc T. Kaufman)
Newsgroups: comp.sys.mac.programmer
Subject: Re: 68000 Assembly question -- overflow
Message-ID: <1991May25.021530.17126@neon.Stanford.EDU>
Date: 25 May 91 02:15:30 GMT
Article-I.D.: neon.1991May25.021530.17126
References: <1991May24.213815.16227@midway.uchicago.edu>
Organization: Computer Science Department, Stanford University, Ca , USA
Lines: 30

In article <1991May24.213815.16227@midway.uchicago.edu> tisu@quads.uchicago.edu (Seth Tisue) writes:
>For class, I'm writing a four-function stack-based calculator in
>assembly language.  The program uses 32-bit signed integers.  My
>question concerns the overflow bit of the status register.  Here's the code:

deleted...

>This code works as expected for all but one case.
>Suppose the two items popped off the stack are $80000001 and $00000001 --
>i.e., in decimal, the function to be performed is (-2147483648) minus one.
>Since that number is the most negative number you can fit into 32 bits,
>it seems to me that this should generate.  However, it does NOT generate
>an overflow (I've stepped through w/MacsBug and checked), and the answer
>returned is $80000000, which is meaningless ("negative zero").

Sigh! What kind of class is this where they don't explain the difference
between 1's complement, 2's complement, and sign-magnitude representations?
The native mode of a 68000 for instance is 2's complement.  In that mode,
$80000001 = -2147483647, and that minus 1 gives $80000000 - -2147483648.
Yes, it is the case that there is no representation in 2's comp form of the
positive number +2147483648 (at least in 32 bits).

In 1's complement and sign-magnitude notation there are, indeed, 2
representations of zero (+0 and -0), and the positive and negative ranges
are symmetric.  But you don't have that, in terms of the 68000 Flag bits
at least.

Marc Kaufman (kaufman@Neon.stanford.edu)
[a believer in the principle that there should be no such thing as an
UNDERGRADUATE CS major]