(unsigned)-INT_MIN

**vippstar@gmail.com** · Jul 21 '08, 09:25 AM

Re: (unsigned)-INT_MIN

On Jul 21, 11:53 am, viza <tom.v...@gm-il.com.obviousc hange.invalid>
wrote:

Hi all
>
int i= INT_MIN;
unsigned int u= -i;
>
Is u guaranteed to have the absolute value of INT_MIN?
>
Why it might not: -i has type (int), and -INT_MIN might be more than
INT_MAX.

So what? corresponding integer types have the same range of values; If
int has N values, unsigned int has N values.
It is guaranteed.
<snip>

**rahul** · Jul 21 '08, 09:25 AM

Re: (unsigned)-INT_MIN

On Jul 21, 1:53 pm, viza <tom.v...@gm-il.com.obviousc hange.invalid>
wrote:

Hi all
>
int i= INT_MIN;
unsigned int u= -i;
>
Is u guaranteed to have the absolute value of INT_MIN?

On my system, u becomes 0 as INT_MIN is -2147483648 and INT_MAX is
2147483647. So, -INT_MIN wraps around to 0.

Why it might not: -i has type (int), and -INT_MIN might be more than
INT_MAX.

-INT_MIN is greater than INT_MAX on most of the systems.

If long is longer than int, I can do -(long)i, but what if we are talking
about the longest integer type available?

What are we looking at? Is it about getting the absolute value of a
signed int in an unsigned int?

**vippstar@gmail.com** · Jul 21 '08, 09:45 AM

Re: (unsigned)-INT_MIN

On Jul 21, 12:21 pm, vipps...@gmail. com wrote:

On Jul 21, 11:53 am, viza <tom.v...@gm-il.com.obviousc hange.invalid>
wrote:Hi all
>

int i= INT_MIN;
unsigned int u= -i;

>

Is u guaranteed to have the absolute value of INT_MIN?

>

Why it might not: -i has type (int), and -INT_MIN might be more than
INT_MAX.

>
So what? corresponding integer types have the same range of values; If
int has N values, unsigned int has N values.
It is guaranteed.
<snip>

I'm sorry, I misunderstood you. You're interested in the case that
INT_MIN < -INT_MAX.
Well, in that case, computing -INT_MIN invokes undefined behavior;
(overflow)
Try this

if(i == INT_MIN)
u += INT_MAX + (unsigned)-(i + INT_MAX);
u = -i;

**Old Wolf** · Jul 21 '08, 09:45 AM

Re: (unsigned)-INT_MIN

On Jul 21, 8:53 pm, viza <tom.v...@gm-il.com.obviousc hange.invalid>
wrote:

int i= INT_MIN;
unsigned int u= -i;
>
Is u guaranteed to have the absolute value of INT_MIN?

No. On a normal (2's complement) system, the
code causes undefined behaviour because the
operation "-i" causes an integer overflow.

Not sure what 'vipps' was on about..

A cast might not work either, because in (unsigned)-i; the cast is too
late, and in -(unsigned)i then i cannot be represented as unsigned.

Signed values can be converted to unsigned ;
(unsigned)i is guaranteed to be UINT_MAX - INT_MIN,
which is INT_MAX on the normal machine.

**vippstar@gmail.com** · Jul 21 '08, 09:55 AM

Re: (unsigned)-INT_MIN

On Jul 21, 12:38 pm, Old Wolf <oldw...@inspir e.net.nzwrote:

On Jul 21, 8:53 pm, viza <tom.v...@gm-il.com.obviousc hange.invalid>
wrote:
>

int i= INT_MIN;
unsigned int u= -i;

>

Is u guaranteed to have the absolute value of INT_MIN?

>
No. On a normal (2's complement) system, the
code causes undefined behaviour because the
operation "-i" causes an integer overflow.
>
Not sure what 'vipps' was on about..

Which one of my two (without counting this one) messages are you
refering to?
I posted a follow-up quoting my first message saying I misunderstood,
and then I gave an answer which I believe to be correct. (and
relevant)

A cast might not work either, because in (unsigned)-i; the cast is too
late, and in -(unsigned)i then i cannot be represented as unsigned.

>
Signed values can be converted to unsigned ;
(unsigned)i is guaranteed to be UINT_MAX - INT_MIN,
which is INT_MAX on the normal machine.

Wrong.
Assuming i is -N with N >= 0, (unsigned)i is UINT_MAX - N + 1.
So (unsigned)INT_M IN would be UINT_MAX + INT_MIN + 1.

**vippstar@gmail.com** · Jul 21 '08, 09:55 AM

Re: (unsigned)-INT_MIN

On Jul 21, 12:36 pm, vipps...@gmail. com wrote:
<snip>

Try this
>
if(i == INT_MIN)
u += INT_MAX + (unsigned)-(i + INT_MAX);
u = -i;

^
insert 'else' there... :-( I should be looking twice at my messages
before posting them.

**viza** · Jul 21 '08, 10:05 AM

Re: (unsigned)-INT_MIN

Hi

On Mon, 21 Jul 2008 08:53:51 +0000, viza wrote:

int i= INT_MIN;
unsigned int u= -i;
>
Is u guaranteed to have the absolute value of INT_MIN?

If long is longer than int, I can do -(long)i, but what if we are
talking about the longest integer type available?

Thanks for the answers so far, but none of them help. Perhaps I can
explain it a bit better:

What I would like is a constant expression of unsigned type for the
absolute value of an constant negative expression of the corresponding
signed type.

Ie:

1) I know i is negative
2) I want (unsigned)abs(i )

The answer simply: ( (unsigned)-i ) for any i except INT_MIN. What is
the answer for INT_MIN? or for any i?

Thanks,

viza

**viza** · Jul 21 '08, 10:05 AM

Re: (unsigned)-INT_MIN

Hi

On Mon, 21 Jul 2008 10:02:32 +0000, viza wrote:

What I would like is a constant expression of unsigned type for the
absolute value of an constant negative expression of the corresponding
signed type.

Thanks for the answers so far, but none of them help

I could use a conditional operation based on vippstars second post, but
is there a neater way?

viza

**Old Wolf** · Jul 21 '08, 11:25 AM

Re: (unsigned)-INT_MIN

On Jul 21, 9:51 pm, vipps...@gmail. com wrote:

On Jul 21, 12:38 pm, Old Wolf <oldw...@inspir e.net.nzwrote:

Not sure what 'vipps' was on about..

>
Which one of my two (without counting this one) messages are you
refering to?

I was referring to your first post on this thread
(and I posted before you posted your correction).

Signed values can be converted to unsigned ;
(unsigned)i is guaranteed to be UINT_MAX - INT_MIN,
which is INT_MAX on the normal machine.

>
Wrong.
Assuming i is -N with N >= 0, (unsigned)i is UINT_MAX - N + 1.
So (unsigned)INT_M IN would be UINT_MAX + INT_MIN + 1.

Knew I should have checked that more before
posting .. the baby was crying though :)

**Old Wolf** · Jul 21 '08, 11:35 AM

Re: (unsigned)-INT_MIN

On Jul 21, 10:02 pm, viza <tom.v...@gm-il.com.obviousc hange.invalid>
wrote:

What I would like is a constant expression of unsigned type for the
absolute value of an constant negative expression of the corresponding
signed type.

-(unsigned)i

Assuming i is negative, then:
(unsigned)i is UINT_MAX + 1 - abs(i)
so the negative of that is
UINT_MAX + 1 - (UINT_MAX + 1 - abs(i))
which is abs(i).

**pete** · Jul 21 '08, 03:05 PM

Re: (unsigned)-INT_MIN

viza wrote:

Hi
>
On Mon, 21 Jul 2008 08:53:51 +0000, viza wrote:
>

>int i= INT_MIN;
>unsigned int u= -i;
>>
>Is u guaranteed to have the absolute value of INT_MIN?

>

>If long is longer than int, I can do -(long)i, but what if we are
>talking about the longest integer type available?

>
Thanks for the answers so far, but none of them help. Perhaps I can
explain it a bit better:
>
What I would like is a constant expression of unsigned type for the
absolute value of an constant negative expression of the corresponding
signed type.
>
Ie:
>
1) I know i is negative
2) I want (unsigned)abs(i )
>
The answer simply: ( (unsigned)-i ) for any i except INT_MIN. What is
the answer for INT_MIN? or for any i?

There is no integer type which is guaranteed
to be able to represent the absolute value of INT_MIN.

--
pete

**vippstar@gmail.com** · Jul 21 '08, 04:15 PM

Re: (unsigned)-INT_MIN

On Jul 21, 5:56 pm, pete <pfil...@mindsp ring.comwrote:
<snip>

There is no integer type which is guaranteed
to be able to represent the absolute value of INT_MIN.

That's incorrect. Do you mean no signed integer type? If so, that is
correct.
unsigned int, unsigned long int, unsigned long long int, all can
represent the absolute value of INT_MIN.

**Keith Thompson** · Jul 21 '08, 05:05 PM

Re: (unsigned)-INT_MIN

vippstar@gmail. com writes:

On Jul 21, 5:56 pm, pete <pfil...@mindsp ring.comwrote:
<snip>

>There is no integer type which is guaranteed
>to be able to represent the absolute value of INT_MIN.

>
That's incorrect. Do you mean no signed integer type? If so, that is
correct.
unsigned int, unsigned long int, unsigned long long int, all can
represent the absolute value of INT_MIN.

That's true on most systems, but it's not guaranteed by the standard.

A conforming system could have padding bits in unsigned int, so that
UINT_MAX == INT_MAX. And unsigned long int and unsigned long long int
could both have the same size and representation as unsigned int
(assuming UINT_MAX >= 2**64-1).

Such a system is unlikely to exist in real life. int and unsigned int
would have to be at least 65 bits (33 bits if you'll settle for C90
conformance).

--
Keith Thompson (The_Other_Keit h) kst-u@mib.org <http://www.ghoti.net/~kst>
Nokia
"We must do something. This is something. Therefore, we must do this."
-- Antony Jay and Jonathan Lynn, "Yes Minister"

**pete** · Jul 21 '08, 09:26 PM

Re: (unsigned)-INT_MIN

vippstar@gmail. com wrote:

On Jul 21, 5:56 pm, pete <pfil...@mindsp ring.comwrote:
<snip>

>There is no integer type which is guaranteed
>to be able to represent the absolute value of INT_MIN.

>
That's incorrect. Do you mean no signed integer type? If so, that is
correct.
unsigned int, unsigned long int, unsigned long long int, all can
represent the absolute value of INT_MIN.

No.
ULONG_MAX can be 4294967295
while INT_MIN can be (-4294967296)
on the same implementation.

--
pete

(unsigned)-INT_MIN

(unsigned)-INT_MIN

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