typecast reminder

[David]

Every once in a while my brain goes dead. in type casting rules, presuming
int is 32 bits, would this always end up 0xFFFFFFFFFFFFFFFF or could it end
up 0x00000000FFFFFFFF:

unsigned __int64 var=(unsigned __int64) -1;

TIA!!

[Hans-Bernhard Bröker]

Assuming __int64 is what it claims, the former. Unsigned arithmetic
works modulo its own cardinality. To get 2^32-1, it would take a
two-step cast:

(unsigned __int64)(unsigned int) -1;

[E. S. Fabian]

Hans-Bernhard Br�ker:

Just as a matter of curiosity, how would these work on a one's complement
machine, or on a sign-and-magnitude machine?
--
Steve

[Hans-Bernhard Bröker]

On 13.05.2011 14:54, E. S. Fabian wrote:
> Hans-Bernhard Bröker:

The same. It doesn't really matter what the representation of signed
numbers is --- the result is defined in terms of values, not representation.

[Mark Wezelenburg]

Hi,

>> --- the result is defined in terms of values, not representation

>> Just as a matter of curiosity, how would these work on a one's complement

For one's complement, the cardinality is 2^N - 1 (rather than 2^N, +/- 0)
so (unsigned)-1 = 2^N-2 (0xfffe) rather than 2^N-1 (0xffff)

Mark

"Hans-Bernhard Bröker" <bro...@physik.rwth-aachen.de> wrote in message
news:iqk543$hgm$1...@www.openwatcom.org...

[Hans-Bernhard Bröker]

On 14.05.2011 00:45, Mark Wezelenburg wrote:
> Hi,
>
>>> --- the result is defined in terms of values, not representation
>>> Just as a matter of curiosity, how would these work on a one's complement
>
> For one's complement, the cardinality is 2^N - 1 (rather than 2^N, +/- 0)

But that only applies to the -1 part of the cast expression. It has
_no_ influence whatsoever on the result of that cast, since that is
performed following the rules for _unsinged_ arithmetic. Signed integer
representation doesn't even enter the picture.

> so (unsigned)-1 = 2^N-2 (0xfffe) rather than 2^N-1 (0xffff)

No.