Re: shame on MISRA

On Fri, 27 Apr 2007 19:55:32 -0400, CBFalconer <cbfalconer@xxxxxxxxx>
wrote:

Dan Henry wrote:
CBFalconer <cbfalconer@xxxxxxxxx> wrote:
Dan Henry wrote:
CBFalconer <cbfalconer@xxxxxxxxx> wrote:

... snip ...

Wrong. C representation may be sign/magnitude or 1's complement.

2's comp: -1 ---> -0x0001 ---> 0xffff
~ 1 ---> ~0x0001 ---> 0xffff (note NOT sign)
~ 0 ---> ~0x0000 ---> 0xffff "
1's comp: -1 ---> -0x0001 ---> 0xffff
~ 1 ---> ~0x0001 ---> 0xfffe (note NOT sign)
~ 0 ---> ~0x0000 ---> 0xffff "
or ~ 0 ---> ~0xffff ---> 0x0000 "
sign mag: -1 ---> -0x0001 ---> 0xffff
~ 1 ---> ~0x0001 ---> 0xfffe (note NOT sign)
~ 0 ---> ~0x0000 ---> 0xffff "

because of the rules for bringing unsigned into range.

~1? I read Marcin to be asking whether ~0 won't set all 1s.

See the additions above. Only -1 always returns 0xffff.

Isn't the integer constant 0 (lacking the unary - operator) in a
1's complement system positive zero, not negative zero? Doesn't
your second ~0 example actually represent ~-0?

In 1's complement -0 is equal to +0.

Right, but I am wondering about the predictability of the underlying
bit patterns.

Are you saying that in 1's complement the integer constant 0 could
have the bit pattern for +0 (all 0's) *or* the bit pattern for -0 (all
1's) and that using the unary - operator on the integer constant 0 can
yield either all 1's or all 0's -- that my memset(&object, 0, sizeof
object) is equally likely to zero all bits as set all bits of the
memory that object occupies?

--
Dan Henry
.

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