A question about taking the absolute value of an integer

[Chad]

I really don't understand how something like..

#define abs(x) (((x) < 0) ? -(x) : (x))

could cause a possible overflow.

[Walter Roberson]

In article <f9************ *************** *******@1g2000h sl.googlegroups .com>,
Chad <cd*****@gmail. comwrote:

>I really don't understand how something like..

>#define abs(x) (((x) < 0) ? -(x) : (x))

>could cause a possible overflow.

e.g., INT_MIN might be -32768.
-(-32768) ---32768 but INT_MAX might be 32767

This is an issue in any implementation that uses twos complement
(unless it reserves the minimum integral value for some reason,
which would be quite unusual.)

It might help to look at the binary values involved.
On a system with a two-byte unsigned short, for example,
SHORT_MIN would be 0x8000 . To take the negative,
toggle each bit and then add 1 to the result. Toggling each
bit of 0x8000 would be 0x7FFF; add 1 and you get 0x8000 which
is the original value back again.
--
"I was very young in those days, but I was also rather dim."
-- Christopher Priest

[Eric Sosman]

Chad wrote:

I really don't understand how something like..

#define abs(x) (((x) < 0) ? -(x) : (x))

could cause a possible overflow.

Find a machine where INT_MIN + INT_MAX == -1 (that's
nearly every machine nowadays) and evaluate abs(INT_MIN).

--
Eric Sosman
es*****@ieee-dot-org.invalid

[Chad]

On Jan 5, 1:10 pm, Eric Sosman <[e-mail] e:

Chad wrote:

I really don't understand how something like..

#define abs(x) (((x) < 0) ? -(x) : (x))

could cause a possible overflow.

Find a machine where INT_MIN + INT_MAX == -1 (that's
nearly every machine nowadays) and evaluate abs(INT_MIN).

--
Eric Sosman
[e-mail]

% more abs.c
#include <stdio.h>
#include <stdlib.h>
#include <limits.h>

int main(void)
{
printf("%u\n",a bs(INT_MIN));
return 0;
}

% ./abs
2147483648

[Coos Haak]

Op Sat, 5 Jan 2008 13:24:09 -0800 (PST) schreef Chad:

On Jan 5, 1:10 pm, Eric Sosman <[e-mail] e:

>Chad wrote:

>>I really don't understand how something like..

>>#define abs(x) (((x) < 0) ? -(x) : (x))

>>could cause a possible overflow.

Find a machine where INT_MIN + INT_MAX == -1 (that's
nearly every machine nowadays) and evaluate abs(INT_MIN).

--
Eric Sosman
[e-mail]

% more abs.c
#include <stdio.h>
#include <stdlib.h>
#include <limits.h>

int main(void)
{
printf("%u\n",a bs(INT_MIN));
return 0;
}

% ./abs
2147483648

Try printf("%n\n" ....
the result will be a negative number. And an absolute value shall be
positive ;-(
--
Coos

[Richard Tobin]

In article <a4************ *************** *******@t1g2000 pra.googlegroup s.com>,
Chad <cd*****@gmail. comwrote:

>% more abs.c
#include <stdio.h>
#include <stdlib.h>
#include <limits.h>

int main(void)
{
printf("%u\n",a bs(INT_MIN));
return 0;
}

% ./abs
2147483648

Perhaps you will see your mistake if you change the printf line
to:

printf("%u %u\n",INT_MIN, abs(INT_MIN));

-- Richard
--
:wq

[David Thompson]

On Sat, 5 Jan 2008 22:33:40 +0100, Coos Haak <ch*****@hccnet .nl>
wrote:

Op Sat, 5 Jan 2008 13:24:09 -0800 (PST) schreef Chad:

printf("%u\n",a bs(INT_MIN));

<snip: appears to work>

Try printf("%n\n" ....
the result will be a negative number. And an absolute value shall be
positive ;-(

%d or %i. %n is something quite different, and wrong here.

- formerly david.thompson1 || achar(64) || worldnet.att.ne t

[Coos Haak]

Op Tue, 22 Jan 2008 05:22:40 GMT schreef David Thompson:

On Sat, 5 Jan 2008 22:33:40 +0100, Coos Haak <ch*****@hccnet .nl>
wrote:

>Op Sat, 5 Jan 2008 13:24:09 -0800 (PST) schreef Chad:

>> printf("%u\n",a bs(INT_MIN));

<snip: appears to work>

>Try printf("%n\n" ....
the result will be a negative number. And an absolute value shall be
positive ;-(

%d or %i. %n is something quite different, and wrong here.

Absolutely, thanks.
--
Coos