unsigned short addition/subtraction overflow

[Andy]

Re: unsigned short addition/subtraction overflow

Richard Heathfield <dontmail@addre ss.co.uk.invali d> wrote in message news:<bsocf3$oo 1$1@sparta.btin ternet.com>...[color=blue]
> Andy wrote:
>[color=green]
> >
> > Can unsigned short be more than 2 bytes long?[/color]
>
> Yes. On a Cray, for example, it might easily be eight bytes long.
>[/color]

Really? From my 1st day at work, I've been taught that
unsigned short is always 2 bytes long while int is variable in length.
You learn something everyday.

Thanks
Andy

[Ben Pfaff]

Re: unsigned short addition/subtraction overflow

bikejog@hotmail .com (Andy) writes:
[color=blue]
> Richard Heathfield <dontmail@addre ss.co.uk.invali d> wrote in message news:<bsocf3$oo 1$1@sparta.btin ternet.com>...[color=green]
> > Andy wrote:
> >[color=darkred]
> > >
> > > Can unsigned short be more than 2 bytes long?[/color]
> >
> > Yes. On a Cray, for example, it might easily be eight bytes long.[/color]
>
> Really? From my 1st day at work, I've been taught that
> unsigned short is always 2 bytes long while int is variable in length.[/color]

Interesting. `unsigned short' is commonly 2 bytes in size, and 2
bytes is commonly 16 bits, but neither is required. 16 bits is
the minimum size of an `unsigned short', though.
--
int main(void){char p[]="ABCDEFGHIJKLM NOPQRSTUVWXYZab cdefghijklmnopq rstuvwxyz.\
\n",*q="kl BIcNBFr.NKEzjwC IxNJC";int i=sizeof p/2;char *strchr();int putchar(\
);while(*q){i+= strchr(p,*q++)-p;if(i>=(int)si zeof p)i-=sizeof p-1;putchar(p[i]\
);}return 0;}

[Richard Heathfield]

Re: unsigned short addition/subtraction overflow

Andy wrote:[color=blue]
> Richard Heathfield wrote:[color=green]
>> Andy wrote:[color=darkred]
>> > Can unsigned short be more than 2 bytes long?[/color]
>>
>> Yes. On a Cray, for example, it might easily be eight bytes long.[/color]
>
> Really? From my 1st day at work, I've been taught that
> unsigned short is always 2 bytes long while int is variable in length.
> You learn something everyday.[/color]

Here's another one for your collection of strange facts. On a 32-bit DSP,
you might easily find that unsigned short is just *one* byte long (that
byte being 32 bits wide, of course).

--
Richard Heathfield : binary@eton.pow ernet.co.uk
"Usenet is a strange place." - Dennis M Ritchie, 29 July 1999.
C FAQ: http://www.eskimo.com/~scs/C-faq/top.html
K&R answers, C books, etc: http://users.powernet.co.uk/eton

[Chris Torek]

Re: unsigned short addition/subtraction overflow

In article <news:aed59298. 0312302033.1943 d7c6@posting.go ogle.com>
Andy <bikejog@hotmai l.com> writes:[color=blue]
>... I guess, then I can say that unsigned short
>addition/subtraction on a machine where sizeof(unsigned short) ==
>sizeof(int) will always yield valid modulus 2^n results. Correct?[/color]

Well, C99 allows for "padding bits" in various representations , in
which case it is possible (though bizarre) to have, e.g.,
CHAR_BIT = 9, sizeof(short) = sizeof(unsigned short) = 2, and
sizeof(int) = 2; yet INT_MAX could be 131071 (no padding bits)
while USHRT_MAX is 65535 (two padding bits). In this case, since
USHRT_MAX < INT_MAX, "unsigned short" promotes to (signed) int
under the ANSI/ISO widening rules!

Such an implementation is certainly bizarre, but it appears to meet
the Standard's requirements.

Note that unless an implementation is *really* strange, we either
have USHRT_MAX > INT_MAX (e.g., 16-bit PDP-11), or 2*USHRT_MAX <=
INT_MAX (e.g., 16-bit "short", 32-bit "int" systems). If the first
is true -- if USHRT_MAX exceeds INT_MAX -- then unsigned short
promotes to unsigned int and arithmetic works mod 2**n. Otherwise,
double USHRT_MAX is still a valid "int" value, so the sum of any
two unsigned short values after they are promoted to signed int
values is a valid, nonnegative "int" value. For instance,
65535 + 65535 is 131070, and on 18-bit-or-more machines this should
be strictly less than INT_MAX (131071 or more). Of course, the
difference (subtraction) of two such values can be a negative "int",
and the sum of three or more such values can overflow.
--
In-Real-Life: Chris Torek, Wind River Systems
Salt Lake City, UT, USA (40°39.22'N, 111°50.29'W) +1 801 277 2603
email: forget about it http://web.torek.net/torek/index.html
Reading email is like searching for food in the garbage, thanks to spammers.

[Kelsey Bjarnason]

Re: unsigned short addition/subtraction overflow

On Tue, 30 Dec 2003 20:37:08 -0800, Andy wrote:
[color=blue]
> Richard Heathfield <dontmail@addre ss.co.uk.invali d> wrote in message news:<bsocf3$oo 1$1@sparta.btin ternet.com>...[color=green]
>> Andy wrote:
>>[color=darkred]
>> >
>> > Can unsigned short be more than 2 bytes long?[/color]
>>
>> Yes. On a Cray, for example, it might easily be eight bytes long.
>>[/color]
>
> Really? From my 1st day at work, I've been taught that
> unsigned short is always 2 bytes long while int is variable in length.
> You learn something everyday.[/color]

Y'all been taught wrong. Types in C and C++ (apart from the new
specified-size types in C99) are defined by minimum required ranges (and
the caveat that some types must be as large or larger than others).

Net result: an unsigned can be one byte, or 128 bytes, as long as it
handles _at least_ the minimum required range.