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From: marv@ISM780.UUCP
Newsgroups: net.arch
Subject: Re: (Orphan) Re: Right shift vs. divide
Message-ID: <27700010@ISM780.UUCP>
Date: Fri, 10-Jan-86 18:34:00 EST
Article-I.D.: ISM780.27700010
Posted: Fri Jan 10 18:34:00 1986
Date-Received: Tue, 14-Jan-86 06:01:11 EST
References: <4772@alice.UUCP.UUCP>
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Nf-ID: #R:alice.UUCP:4772:ISM780:27700010:000:1340
Nf-From: ISM780!marv    Jan 10 18:34:00 1986


>> 1)if the operand is positive right shift.
>>
>> 2)if the operand is negative, negate the operand before shifting, then
>> right shift, and then negate the result.
>  ^^^^^^^^^^^
>>
>> if you want, try the infamous right shift all 1's (-1).
>>
>> the division is round toward zero (ala fortran).
>>
>> the MV series from data general perform arithmetic right shifting
>> correctly

>Still more evidence that the problem is harder than it looks:
>the above algorithm fails if the dividend is the most negative number.
>Consider, for example, a 32-bit machine.  Let's divide
>-2147483648 by 2.  The answer should be -1073741824.

>        -2147483648 is 80000000 hex.  Negate it and you
>        get integer overflow.  If this is masked, most machines
>        will give you 80000000 hex back.
>
>        Shift 80000000 right 1 bit, giving C0000000.
>
>        Negate this again, giving 40000000.
>
>Thus, the sign gets lost.
>/* End of text from ISM780:net.arch */
>

Boy this shure is harder than it looks. Note that the algorithm said "shift
right" *not* "arithmetic shift right"

    800000000>>1 is 40000000 (not C0000000)

    40000000 negated is C0000000 which is -1073741824

However I agree with Tim. It would be better if -1/2 produced -1. If you
are interested in my reasons, send mail.

	 Marv Rubinstein -- Interactive Systems