From: Shalom Bresticker (Shalom.Bresticker@motorola.com)
Date: Sun May 19 2002 - 04:09:25 PDT
Two comments, see below.
Shalom
Steven Sharp wrote:
> >Sec 4.1.12 says that a 0 or a signbit is filled in based on the signedness of
> the
> >result type. Later it says "The result signedness is .... and the remainder of
> the
> >expression as outlined in sec 4.5.1". However, I do not see the arith shift
> operators
> >listed in Table 29.
>
> They should be in the same entry as the other shifts. I believe this
> erratum was reported but not fixed before printing.
Yes, it is issue 18 in the database.
> >I would have thought that whether a 0 or a signbit is filled based solely on
> >the signedness of the first operand and nothing else.
>
> So will many other people. This is likely to lead to much misunderstanding
> of the behavior of signed arithmetic, but that is how it is defined. The
> behavior of sign-conversion matches the behavior of width-conversion, where
> all operands are converted before any operators are applied. This makes all
> conversions consistent within the language. The only problem is that most
> users don't know how the width conversion works to start with. It was
> designed to work the way it does precisely so they could get away with
> not knowing how it works and still get the expected results. The same is
> not true when the same conversion order is applied to signedness, but that
> is how it is.
Question on "The
behavior of sign-conversion matches the behavior of width-conversion":
As far as I can see, an expression width is determined also by its LHS whereas
its signedness is not. True?
Thanks,
Shalom
-- Shalom Bresticker Shalom.Bresticker@motorola.com Design & Reuse Methodology Tel: +972 9 9522268 Motorola Semiconductor Israel, Ltd. Fax: +972 9 9522890 POB 2208, Herzlia 46120, ISRAEL Cell: +972 50 441478
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