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int

if
{
unsigned u = -INT_MIN;
int i, sign = -1;
i = sign * u;

then i == INT_MIN or not?
Nov 14 '05 #1
4 1261

"RoSsIaCrIiLoIA" <n@esiste.ee> wrote in message
if
{
unsigned u = -INT_MIN;
int i, sign = -1;
i = sign * u;

then i == INT_MIN or not?

The calculation is allowed to overflow, producing implementation-defined (I
think) behaviour.
On most platforms, INT_MIN = -INT_MAX - 1. Taking the negative produces
INT_MAX + 1, which will overflow to INT_MIN. Multiplying by -1 will further
overflow to INT_MIN, for by a contorted route you will actually get the
answer you suggest.
Nov 14 '05 #2
"RoSsIaCrIiLoIA" <n@esiste.ee> wrote in message
news:mb********************************@4ax.com...
if
{
unsigned u = -INT_MIN;
-INT_MIN is performed using signed int arithmetic (before the assignment), and it can
overflow (undefined behaviour).

Note that even if you do...

unsigned u = INT_MIN;
u = -u;

You have no guarantee that u will be the magnitude of INT_MIN. Since UINT_MAX can equal
INT_MAX, you might have u == 0.
int i, sign = -1;
i = sign * u;

then i == INT_MIN or not?


sign * u will be performed in unsigned int arithmetic, so the result will always be
positive. The conversion on assignment to i will be implementation defined if the value is
outside the range [0..INT_MAX]. So, it could theoretically set i to anything. C99 even
allows an implementation defined signal to be raised.

Even C99 (optional) intN_t types, whilst necessarily unpadded twos complement integers,
are not immune from implementation defined conversions and potential signals if assigned a
value that is outside their range.

--
Peter
Nov 14 '05 #3
On Sun, 2 May 2004 13:35:05 +1000, "Peter Nilsson" <ai***@acay.com.au>
wrote:
"RoSsIaCrIiLoIA" <n@esiste.ee> wrote in message
news:mb********************************@4ax.com.. .
if
{
unsigned u = -INT_MIN;


-INT_MIN is performed using signed int arithmetic (before the assignment), and it can
overflow (undefined behaviour).

Note that even if you do...

unsigned u = INT_MIN;
u = -u;


what do you say for
unsigned u = -(INT_MIN + 1);

u += 1;
?
Thanks
Nov 14 '05 #4
RoSsIaCrIiLoIA <n@esiste.ee> wrote in message news:<vt********************************@4ax.com>. ..
On Sun, 2 May 2004 13:35:05 +1000, "Peter Nilsson" <ai***@acay.com.au>
wrote:
"RoSsIaCrIiLoIA" <n@esiste.ee> wrote in message
news:mb********************************@4ax.com...
if
{
unsigned u = -INT_MIN;


-INT_MIN is performed using signed int arithmetic (before the assignment),
and it can overflow (undefined behaviour).

Note that even if you do...

unsigned u = INT_MIN;
u = -u;


what do you say for
unsigned u = -(INT_MIN + 1);
u += 1;


Or just: u = 0u - INT_MIN;

This is well defined [although one comp.std.c regular might disagree about
-(INT_MIN+1)] however...

This can still leave you with u == 0 on theoretical twos complement machines
where UINT_MAX == INT_MAX.

--
Peter
Nov 14 '05 #5

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