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How to extract bytes from long?

P: n/a
RB
How to extract bytes from long, starting from the last byte?
For example, I have a long number:
0x12345678
I need to represent it as the following bytes list:
0x78, 0x56, 0x34, 0x12

Thanks in advance,
Rita
Nov 13 '05 #1
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44 Replies


P: n/a
RB wrote:
How to extract bytes from long, starting from the last byte?
For example, I have a long number:
0x12345678
I need to represent it as the following bytes list:
0x78, 0x56, 0x34, 0x12

Thanks in advance,
Rita


Look at bitwise operators. &, |, ^, <<, >>. Then look up bitmasks
and how to use them and everything should be rather straightforward.

--
Thomas.
"What is the future c existence which does not do in all languages"

Nov 13 '05 #2

P: n/a
On Wed, 15 Oct 2003 01:59:04 -0700, RB wrote:
How to extract bytes from long, starting from the last byte?


#include <stdio.h>

int main() {

unsigned long value = 0x12345678;
int i;

printf("%#lx\n",value);

for (i = sizeof value; i > 0; --i) {
printf("%#x\n", value & 0xff);
value >>= 8;
}
return 0;
}

--
NPV

"the large print giveth, and the small print taketh away"
Tom Waits - Step right up

Nov 13 '05 #3

P: n/a


RB wrote:
How to extract bytes from long, starting from the last byte?
For example, I have a long number:
0x12345678
I need to represent it as the following bytes list:
0x78, 0x56, 0x34, 0x12

Thanks in advance,
Rita


Sounds a bit homework-y, but as a hint try this:

#include <stdio.h>

int main(void)
{
long val=0x12345678;

printf("0x%x\n",val >> 16 & 0xff );

return 0;
}

Regards,

Ed.

Nov 13 '05 #4

P: n/a
Nils Petter Vaskinn wrote:

On Wed, 15 Oct 2003 01:59:04 -0700, RB wrote:
How to extract bytes from long, starting from the last byte?


#include <limits.h> #include <stdio.h>

int main() {

unsigned long value = 0x12345678;
int i;

printf("%#lx\n",value);

for (i = sizeof value; i > 0; --i) {
printf("%#x\n", value & 0xff);
value >>= 8;
/*
** You realise that you don't know the size of value,
** so you might as well go all the way.
*/
printf("%#x\n", value & (unsigned char)-1);
value >>= CHAR_BIT;
}
return 0;
}


--
pete
Nov 13 '05 #5

P: n/a
pete wrote:

Nils Petter Vaskinn wrote:

On Wed, 15 Oct 2003 01:59:04 -0700, RB wrote:
How to extract bytes from long, starting from the last byte?


.... and now for the (sizeof(long)==1) portable way:

/* BEGIN new.c */

#include <stdio.h>
#include <limits.h>

int main(void)
{
unsigned long value = 0x12345678;
size_t i;

printf("%#lx\n",value);
printf("%#x\n", value & (unsigned char)-1);
for (i = sizeof value - 1; i != 0; --i) {
value >>= CHAR_BIT;
printf("%#x\n", value & (unsigned char)-1);
}
return 0;
}

/* END new.c */

and now for the way which interprets "last byte"
as being the one furthest from ((char*)&value)

/* BEGIN new2.c */

#include <stdio.h>

int main(void)
{
unsigned long value = 0x12345678;
unsigned char *pointer = (unsigned char*)&value + sizeof value;

printf("%#lx\n",value);
do {
--pointer;
printf("%#x\n", (unsigned)*pointer);
} while (pointer != (unsigned char*)&value);
return 0;
}

/* END new2.c */

--
pete
Nov 13 '05 #6

P: n/a
ri********@yahoo.com (RB) wrote in message news:<3a*************************@posting.google.c om>...
How to extract bytes from long, starting from the last byte?
For example, I have a long number:
0x12345678
I need to represent it as the following bytes list:
0x78, 0x56, 0x34, 0x12

Thanks in advance,
Rita


There are a couple of ways. The safest is to use a bitmask (0xFF),
the bitwise & operator, and the >> and << shift operators.
Alternately, you can treat the long as an array of unsigned char by
creating an unsigned char pointer and setting it to the same address
as the long (unsigned char *p = (unsigned char *) &mylong;) and then
either use array subscript notation to access individual bytes or
"walk" the array by incrementing p, but you have to be aware of
endianness issues (i.e., on a little-endian machine, p[0] would be the
LSB, whereas on a big-endian machine it would be the MSB). The first
method (bitmask and shift) works the same regardless of endian issues.
Nov 13 '05 #7

P: n/a
pete <pf*****@mindspring.com> wrote in message news:<3F***********@mindspring.com>...
pete wrote:

Nils Petter Vaskinn wrote:

On Wed, 15 Oct 2003 01:59:04 -0700, RB wrote:

> How to extract bytes from long, starting from the last byte?

... and now for the (sizeof(long)==1) portable way:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
/* BEGIN new.c */

#include <stdio.h>
#include <limits.h>

int main(void)
{
unsigned long value = 0x12345678;
size_t i;

printf("%#lx\n",value);
printf("%#x\n", value & (unsigned char)-1);
for (i = sizeof value - 1; i != 0; --i) {
value >>= CHAR_BIT;
UB if sizeof(long) == 1.

value = value >> (CHAR_BIT - 1) >> 1;
printf("%#x\n", value & (unsigned char)-1);
}
return 0;
}

/* END new.c */


--
Peter
Nov 13 '05 #8

P: n/a
pete wrote:
Nils Petter Vaskinn wrote:
On Wed, 15 Oct 2003 01:59:04 -0700, RB wrote:
How to extract bytes from long, starting from the last byte?


#include <limits.h>
#include <stdio.h>

int main() {

unsigned long value = 0x12345678;
int i;

printf("%#lx\n",value);

for (i = sizeof value; i > 0; --i) {
printf("%#x\n", value & 0xff);
value >>= 8;


/*
** You realise that you don't know the size of value,
** so you might as well go all the way.
*/
printf("%#x\n", value & (unsigned char)-1);
value >>= CHAR_BIT;
}
return 0;
}


You need neither CHAR_BIT nor shifts nor limits.h nor sizeof:

for (i = 8; i > 0; --i) {
printf("%x ", value % 256);
value /= 256;
}
putchar('\n'); /* <--AND HERE is where the \n goes */
return 0;
}

and the result is portable.

--
Chuck F (cb********@yahoo.com) (cb********@worldnet.att.net)
Available for consulting/temporary embedded and systems.
<http://cbfalconer.home.att.net> USE worldnet address!
Nov 13 '05 #9

P: n/a
pete <pf*****@mindspring.com> wrote in message news:<3F***********@mindspring.com>...
printf("%#x\n", value & (unsigned char)-1);
for (i = sizeof value - 1; i != 0; --i) {
value >>= CHAR_BIT;
printf("%#x\n", value & (unsigned char)-1);
}
return 0;
}


I would use ~0 for "all 1s" rather than -1.

Sam
Nov 13 '05 #10

P: n/a
op*****@yahoo.com (Samuel Barber) wrote in message news:<37**************************@posting.google. com>...
pete <pf*****@mindspring.com> wrote in message news:<3F***********@mindspring.com>...
printf("%#x\n", value & (unsigned char)-1);
for (i = sizeof value - 1; i != 0; --i) {
value >>= CHAR_BIT;
printf("%#x\n", value & (unsigned char)-1);
}
return 0;
}


I would use ~0 for "all 1s" rather than -1.


That risks a trap representation under C99.

--
Peter
Nov 13 '05 #11

P: n/a
Peter Nilsson wrote:

pete <pf*****@mindspring.com> wrote in message news:<3F***********@mindspring.com>...
pete wrote:

Nils Petter Vaskinn wrote:
>
> On Wed, 15 Oct 2003 01:59:04 -0700, RB wrote:
>
> > How to extract bytes from long, starting from the last byte?


... and now for the (sizeof(long)==1) portable way:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

/* BEGIN new.c */

#include <stdio.h>
#include <limits.h>

int main(void)
{
unsigned long value = 0x12345678;
size_t i;

printf("%#lx\n",value);
printf("%#x\n", value & (unsigned char)-1);
for (i = sizeof value - 1; i != 0; --i) {
value >>= CHAR_BIT;


UB if sizeof(long) == 1.


No.
If sizeof(long) == 1,
then the loop doesn't execute.
printf("%#x\n", value & (unsigned char)-1);
}
return 0;
}

/* END new.c */


--
pete
Nov 13 '05 #12

P: n/a
ai***@acay.com.au (Peter Nilsson) wrote in message news:<63**************************@posting.google. com>...
op*****@yahoo.com (Samuel Barber) wrote in message news:<37**************************@posting.google. com>...
pete <pf*****@mindspring.com> wrote in message news:<3F***********@mindspring.com>...
printf("%#x\n", value & (unsigned char)-1);
for (i = sizeof value - 1; i != 0; --i) {
value >>= CHAR_BIT;
printf("%#x\n", value & (unsigned char)-1);
}
return 0;
}


I would use ~0 for "all 1s" rather than -1.


That risks a trap representation under C99.


Nonsense. ~0 is idiomatic C. Aren't you worried about the "risk" that
-1 may not be implemented as all-1s? That is, after all, an
implementation detail. It's not true for sign-magnitude or 1's
complement, for example.

Sam
Nov 13 '05 #13

P: n/a
CBFalconer wrote:

pete wrote:
Nils Petter Vaskinn wrote:
On Wed, 15 Oct 2003 01:59:04 -0700, RB wrote:

> How to extract bytes from long, starting from the last byte?


#include <limits.h>
#include <stdio.h>

int main() {

unsigned long value = 0x12345678;
int i;

printf("%#lx\n",value);

for (i = sizeof value; i > 0; --i) {
printf("%#x\n", value & 0xff);
value >>= 8;


/*
** You realise that you don't know the size of value,
** so you might as well go all the way.
*/
printf("%#x\n", value & (unsigned char)-1);
value >>= CHAR_BIT;
}
return 0;
}


You need neither CHAR_BIT nor shifts nor limits.h nor sizeof:

for (i = 8; i > 0; --i) {
printf("%x ", value % 256);
value /= 256;
}
putchar('\n'); /* <--AND HERE is where the \n goes */
return 0;
}

and the result is portable.


I was addressing the more general subject,
in the subject line of this thread: "How to extract bytes from long?",
rather than how to extract bytes from 0x12345678
or any other number which doesn't require more than 32 bits.

--
pete
Nov 13 '05 #14

P: n/a
Samuel Barber wrote:

ai***@acay.com.au (Peter Nilsson) wrote in message news:<63**************************@posting.google. com>...
op*****@yahoo.com (Samuel Barber) wrote in message news:<37**************************@posting.google. com>...
pete <pf*****@mindspring.com> wrote in message news:<3F***********@mindspring.com>...
> printf("%#x\n", value & (unsigned char)-1);
> for (i = sizeof value - 1; i != 0; --i) {
> value >>= CHAR_BIT;
> printf("%#x\n", value & (unsigned char)-1);
> }
> return 0;
> }

I would use ~0 for "all 1s" rather than -1.


That risks a trap representation under C99.


Nonsense. ~0 is idiomatic C. Aren't you worried about the "risk" that
-1 may not be implemented as all-1s? That is, after all, an
implementation detail. It's not true for sign-magnitude or 1's
complement, for example.


You're wrong about everything.

~0 is negative zero in ones complement.
Implementations are allowed trap negative zero.

-1 is a value, not a bit pattern.
The value of negative one, cast to unsigned char, is UCHAR_MAX.

--
pete
Nov 13 '05 #15

P: n/a
On Fri, 17 Oct 2003 02:44:03 +0000, pete wrote:
Samuel Barber wrote:

ai***@acay.com.au (Peter Nilsson) wrote in message news:<63**************************@posting.google. com>...
> op*****@yahoo.com (Samuel Barber) wrote in message news:<37**************************@posting.google. com>...
> > pete <pf*****@mindspring.com> wrote in message news:<3F***********@mindspring.com>...
> > > printf("%#x\n", value & (unsigned char)-1);
> > > for (i = sizeof value - 1; i != 0; --i) {
> > > value >>= CHAR_BIT;
> > > printf("%#x\n", value & (unsigned char)-1);
> > > }
> > > return 0;
> > > }
> >
> > I would use ~0 for "all 1s" rather than -1.
>
> That risks a trap representation under C99.
Nonsense. ~0 is idiomatic C. Aren't you worried about the "risk" that
-1 may not be implemented as all-1s? That is, after all, an
implementation detail. It's not true for sign-magnitude or 1's
complement, for example.


You're wrong about everything.

~0 is negative zero in ones complement.
Implementations are allowed trap negative zero.


~0u cannot be a trap representation, however.

6.5.3.3
4 The result of the ~ operator is the bitwise complement of its
(promoted) operand ... If the promoted type is an unsigned type,
the expression ~E is equivalent to the maximum value representable
in that type minus E.
-1 is a value, not a bit pattern.
The value of negative one, cast to unsigned char, is UCHAR_MAX.


Nov 13 '05 #16

P: n/a
pete <pf*****@mindspring.com> wrote in message news:<3F***********@mindspring.com>...
Samuel Barber wrote:

ai***@acay.com.au (Peter Nilsson) wrote in message news:<63**************************@posting.google. com>...
op*****@yahoo.com (Samuel Barber) wrote in message news:<37**************************@posting.google. com>...
> pete <pf*****@mindspring.com> wrote in message news:<3F***********@mindspring.com>...
> > printf("%#x\n", value & (unsigned char)-1);
> > for (i = sizeof value - 1; i != 0; --i) {
> > value >>= CHAR_BIT;
> > printf("%#x\n", value & (unsigned char)-1);
> > }
> > return 0;
> > }
>
> I would use ~0 for "all 1s" rather than -1.

That risks a trap representation under C99.
Nonsense. ~0 is idiomatic C. Aren't you worried about the "risk" that
-1 may not be implemented as all-1s? That is, after all, an
implementation detail. It's not true for sign-magnitude or 1's
complement, for example.


You're wrong about everything.

~0 is negative zero in ones complement.
Implementations are allowed trap negative zero.


You're just repeating words, without any understanding. There's no
such thing as "negative zero" (or negative anything) in the context of
bitwise operations. How can a bitwise operation trap? It can't.
-1 is a value, not a bit pattern.
The value of negative one, cast to unsigned char, is UCHAR_MAX.


Hello? The point is that -1 is ***being used as*** a bit pattern. The
intent is to get "all 1s", which is true if the integer representation
is 2's complement; that's an implicit assumption of the code. (This is
the best reason not to use -1: the intent is not perfectly clear).

Sam
Nov 13 '05 #17

P: n/a

On Fri, 16 Oct 2003, Samuel Barber wrote:

pete wrote in message news:<3F***********@mindspring.com>...
Samuel Barber wrote:
(Peter Nilsson) wrote...
> Samuel Barber wrote...
> > pete wrote in message news:<3F***********@mindspring.com>...
> > >
> > > printf("%#x\n", value & (unsigned char)-1);
> >
> > I would use ~0 for "all 1s" rather than -1.
>
> That risks a trap representation under C99.

Nonsense. ~0 is idiomatic C. Aren't you worried about the "risk" that
-1 may not be implemented as all-1s? That is, after all, an
implementation detail. It's not true for sign-magnitude or 1's
complement, for example.
You're wrong about everything.

~0 is negative zero in ones complement.
Implementations are allowed trap negative zero.


You're just repeating words, without any understanding. There's no
such thing as "negative zero" (or negative anything) in the context of
bitwise operations.


It is AFAIK implementation-defined whether the concept of "negative
zero" is meaningful in standard C (depending on the representation
of signed integers).
How can a bitwise operation trap? It can't.
Of course it can! (Why wouldn't it? And do modern digital computers
perform any operations that *aren't* bitwise, anyway?)
-1 is a value, not a bit pattern.
The value of negative one, cast to unsigned char, is UCHAR_MAX.


To clarify pete's point:

(uchar)-1 == (uchar)-((int)1) == (-(int)1)+UCHAR_MAX == UCHAR_MAX-1

....which is guaranteed to have an all-ones bit pattern, regardless
of padding or n's-complement.

Hello? The point is that -1 is ***being used as*** a bit pattern.
Then it's being used incorrectly. -1 is *not* a bit pattern, it's
an integer expression equal to the additive inverse of the integer 1.
It doesn't have a "bit pattern" per se.
The intent is to get "all 1s", which is true if the integer
representation is 2's complement; that's an implicit assumption
of the code.
If that *were* an implicit assumption of the code, then the code
would be broken. But it's not. Pete's code AFAICT doesn't assume
anything earth-shattering about the integer representations used
by the target system.
(This is the best reason not to use -1: the intent is not
perfectly clear).


However, it *does* produce the right answer, which is a point in
its favor. ~0 might trap, and in any case I think (unsigned char)-1
has a bit more aesthetic value to it (YMMV, of course).

And if you want everything to be *perfectly* clear, then you might
want to consider a different programming language. C is just
full of 'for (i=0; a[i]; ++i)'s and 'while (*s++ = *t++)'s; if
a simple -1 throws you, then you're in trouble. :-)

-Arthur
Nov 13 '05 #18

P: n/a
On Thu, 16 Oct 2003 23:53:05 -0700, Samuel Barber wrote:
pete <pf*****@mindspring.com> wrote in message news:<3F***********@mindspring.com>...
Samuel Barber wrote:
>
> ai***@acay.com.au (Peter Nilsson) wrote in message news:<63**************************@posting.google. com>...
> > op*****@yahoo.com (Samuel Barber) wrote in message news:<37**************************@posting.google. com>...
> > > pete <pf*****@mindspring.com> wrote in message news:<3F***********@mindspring.com>...
> > > > printf("%#x\n", value & (unsigned char)-1);
> > > > for (i = sizeof value - 1; i != 0; --i) {
> > > > value >>= CHAR_BIT;
> > > > printf("%#x\n", value & (unsigned char)-1);
> > > > }
> > > > return 0;
> > > > }
> > >
> > > I would use ~0 for "all 1s" rather than -1.
> >
> > That risks a trap representation under C99.
>
> Nonsense. ~0 is idiomatic C. Aren't you worried about the "risk" that
> -1 may not be implemented as all-1s? That is, after all, an
> implementation detail. It's not true for sign-magnitude or 1's
> complement, for example.


You're wrong about everything.

~0 is negative zero in ones complement.
Implementations are allowed trap negative zero.


You're just repeating words, without any understanding. There's no
such thing as "negative zero" (or negative anything) in the context of
bitwise operations. How can a bitwise operation trap? It can't.


6.2.6.2 Integer types
2 ... (It is implementation-defined) whether the value with ...
sign bit 1 and all value bits 1 (for one's complement), is a trap
representation or a normal value. In the case of ... one's
complement, if this representation is a normal value it is called
a negative zero.
...
3 If the implementation supports negative zeros, they shall be
generated only by: the &, |, ^, ~, <<, and >> operators with
arguments that produce such a value;
...
4 If the implementation does not support negative zeros, the
behavior of the &, |, ^, ~, <<, and >> operators with arguments
that would produce such a value is undefined.
Nov 13 '05 #19

P: n/a
Arthur J. O'Dwyer wrote:
....
(uchar)-1 == (uchar)-((int)1) == (-(int)1)+UCHAR_MAX == UCHAR_MAX-1


You must have miscalculated somewhere. (uc)-1 should be UCHAR_MAX.

(uc)-1 == (uc)(0 - 1) == (uc)(UINT_MAX + 1 - 1) == (uc)(UINT_MAX) == UCHAR_MAX

Jirka

Nov 13 '05 #20

P: n/a
Jirka Klaue wrote:
....
(uc)-1 == (uc)(0 - 1) == (uc)(UINT_MAX + 1 - 1) == (uc)(UINT_MAX) == UCHAR_MAX


s/UI/I/g

Jirka

Nov 13 '05 #21

P: n/a
op*****@yahoo.com (Samuel Barber) wrote in message news:<37**************************@posting.google. com>...
pete <pf*****@mindspring.com> wrote in message news:<3F***********@mindspring.com>...
-1 is a value, not a bit pattern.
The value of negative one, cast to unsigned char, is UCHAR_MAX.


Hello? The point is that -1 is ***being used as*** a bit pattern. The
intent is to get "all 1s", which is true if the integer representation
is 2's complement; that's an implicit assumption of the code. (This is
the best reason not to use -1: the intent is not perfectly clear).


(Quoting myself)

Please disregard this part of my reply. I misinterpreted what Pete was saying.

Sam
Nov 13 '05 #22

P: n/a
"Arthur J. O'Dwyer" <aj*@nospam.andrew.cmu.edu> wrote in message news:<Pi**********************************@unix48. andrew.cmu.edu>...
Samuel Barber wrote:

How can a bitwise operation trap? It can't.


Of course it can! (Why wouldn't it? And do modern digital computers
perform any operations that *aren't* bitwise, anyway?)


The C bitwise operators are &, |, ^, and ~ (>> and << are also
included in this catagory, although bitwise is a misnomer in the case
of shifts). "Trap" in the context of this discussion seems to mean
"detect an illegal value"; well, there are only two values of interest
to bitwise operators (0 and 1), and they are both legal. So how can it
trap?

Sam
Nov 13 '05 #23

P: n/a
Sheldon Simms <sh**********@yahoo.com> wrote in message news:<pa****************************@yahoo.com>...
6.2.6.2 Integer types
2 ... (It is implementation-defined) whether the value with ...
sign bit 1 and all value bits 1 (for one's complement), is a trap
representation or a normal value. In the case of ... one's
complement, if this representation is a normal value it is called
a negative zero.
...
3 If the implementation supports negative zeros, they shall be
generated only by: the &, |, ^, ~, <<, and >> operators with
arguments that produce such a value;
...
4 If the implementation does not support negative zeros, the
behavior of the &, |, ^, ~, <<, and >> operators with arguments
that would produce such a value is undefined.


Okay, but if we are to believe this hocus pocus, there is no way to
avoid the hypothetical trapping. It makes no difference whether you
use (unsigned char)-1, (unsigned char)~0, or UCHAR_MAX, since they all
evaluate to the same thing. All are equally right or equally wrong.

Sam
Nov 13 '05 #24

P: n/a
op*****@yahoo.com (Samuel Barber) wrote:
"Arthur J. O'Dwyer" <aj*@nospam.andrew.cmu.edu> wrote in message news:<Pi**********************************@unix48. andrew.cmu.edu>...
> > Samuel Barber wrote:
> How can a bitwise operation trap? It can't.
Of course it can! (Why wouldn't it? And do modern digital computers
perform any operations that *aren't* bitwise, anyway?)


The C bitwise operators are &, |, ^, and ~ (>> and << are also
included in this catagory, although bitwise is a misnomer in the case
of shifts).


Not from the standard's POV. In fact, they are referred to as "bitwise
shift operators" explicitly.
"Trap" in the context of this discussion seems to mean
"detect an illegal value"; well, there are only two values of interest
to bitwise operators (0 and 1), and they are both legal. So how can it
trap?


Read C99 6.2.6.2, Sheldon Simms already quoted the relevant parts in
reply to your other post.

Regards
--
Irrwahn
(ir*******@freenet.de)
Nov 13 '05 #25

P: n/a
>Sheldon Simms <sh**********@yahoo.com> wrote in message news:<pa****************************@yahoo.com>...
6.2.6.2 Integer types
2 ... (It is implementation-defined) whether the value with ...
sign bit 1 and all value bits 1 (for one's complement), is a trap
representation or a normal value. In the case of ... one's
complement, if this representation is a normal value it is called
a negative zero. ...

In article <37**************************@posting.google.com >,
Samuel Barber <op*****@yahoo.com> wrote:Okay, but if we are to believe this hocus pocus, there is no way to
avoid the hypothetical trapping. It makes no difference whether you
use (unsigned char)-1, (unsigned char)~0, or UCHAR_MAX, since they all
evaluate to the same thing. All are equally right or equally wrong.


You appear to have an incorrect "mental model" of how C works.
This is not surprising; I suspect most people do. (One needs to
have worked with those oddball ones'-complement machines to really
have a feel for this stuff.)

The language is not defined in terms of "what happens on a PDP-11",
nor "what happens on a VAX", nor even "what happens on an x86 or
other CPU produced within the last few years". Rather, it is defined
in terms of an "abstract machine". A C compiler writer must map
from "abstract machine" to "real machine" in some way.

The section quoted above (along with others) define how the abtract
machine is to work. In the abstract machine, writing:

~0

means:

- make an int with the value 0
- now, flip all the bits

This process *can* give rise to a "trap representation" on a ones'
complement machine.

On the other hand, writing:

-1

means:

- make an int with the value 1
- now, negate it

This process *must* produce the (ordinary signed int) value -1.
On a ones' complement machine, this value in binary is a sequence
of 1 bits followed by a zero, e.g., 111111111111111110 -- 17 1 bits
and then a 0 -- on an 18-bit-int ones' complement CPU. (The CPU
I am using as a model here is the Univac 11xx, which has 9, 18,
and 36 bit integers and *does* use ones' complement.)

Converting any ordinary signed int to type "unsigned char" *must*
produce a valid unsigned char bit pattern and value -- these are
defined as more or less the same thing in the abstract machine --
and the process by which a negative signed int is transformed into
a (positive) unsigned char is defined mathematically. If the
signed int has value -1, the result must be UCHAR_MAX, which is
a valid bit pattern that consists of all-1-bits, e.g., 111111111
(9 ones) on a 9-bit-byte ones' complement CPU.

At this point you are probably ready to hit your "post follow-up"
key or mouseable button or whatnot, saying: "What?! HOLD ON! JUST
A CONSARNED MINUTE! That all-1-bits pattern, you just said it's
a trap representation, now you say it's a valid value?!?" Yep.
How can it be both?

The answer lies in the *type* of the value. When the *type* of
the value is "signed int", an all-one-bits pattern is allowed to
be a "trap representation". When the type is "unsigned int", this
is *not* allowed. If the target CPU makes this a royal pain in
the butt, well, too bad for the C compiler implementor and/or user
-- "unsigned"s are going to be difficult and/or slow. But if you
*need* all-one-bits patterns, you -- as a C programmer -- should
use "unsigned" arithmetic, which is well-behaved and avoids all
these "trap representation" things. Moreover, given:

unsigned int ui = UINT_MAX;

the sequence:

ui++;

is *guaranteed* to cause ui to "roll over" to zero, without trapping
at runtime with an overflow error. With ordinary signed ints there
is no guarantee -- they may "roll over" (from positive to negative
or vice versa) or they may trap at runtime, whichever the implementor
finds easier or "better".

At the edges, the rules for C can get pretty complicated, but
there *are* simple answers for the common cases:

- If you need an ordinary signed integer and do not believe
you will overflow it, use an ordinary signed integer. (Use
"long" if your range is -2 billion to +2 billion; in C99, use
"long long" if your range is -9 quintillion to +9 quintillion.
Numerically these are 2147483647 and 9223372036854775807
respectively, in case you-the-reader are someone who uses
"milliard". :-) Ordinary "int" is only guaranteed to handle
[-32767..+32767], even though it often handles the 2 billion
number.)

- If you need modular "clock arithmetic", use an unsigned integer.

- If you need to do bitwise operations, use an unsigned integer.

- If you need exact, precisely defined behavior in *all* cases,
use an unsigned integer, synthesizing your own signed values
from these if desired. (In other words, build your own ones'
or two's complement or sign-and-magnitude system.)

Incidentally, one trick proposed (but not actually used on the
Univac) for unsigned integers vs. trap representations is, e.g.,
to have "unsigned int" be only 17 bits, while ordinary signed int
is 18 bits. Then UINT_MAX and INT_MAX are the same number (!),
and "unsigned"ness is achieved mainly by forcing the sign bit to
stay off. This appears to be allowed by the C standard. It is
therefore possible that the "simple rules" *still* do not achieve
the desired effect, depending on what that desired effect might
be.
--
In-Real-Life: Chris Torek, Wind River Systems
Salt Lake City, UT, USA (4039.22'N, 11150.29'W) +1 801 277 2603
email: forget about it http://67.40.109.61/torek/index.html (for the moment)
Reading email is like searching for food in the garbage, thanks to spammers.
Nov 13 '05 #26

P: n/a
"Arthur J. O'Dwyer" <aj*@nospam.andrew.cmu.edu> schrieb im Newsbeitrag
news:Pi**********************************@unix48.a ndrew.cmu.edu...
However, it *does* produce the right answer, which is a point in
its favor. ~0 might trap, and in any case I think (unsigned char)-1
has a bit more aesthetic value to it (YMMV, of course).


Why should ~0 trap??? it results in the 1's complement of 0 which means all
bits are 1's. Padding bits are not affected by this operation, however the
values of padding bits should never be of your interest.

A Trap representation can *only* be generated when manipulating the value
using pointers which aren't the type of the value or doesn't start at the
exact address of the value. Wrong usage of an union can also result in a
Trap representation.

But all arithmetic or bitwise operations cannot result in a
trap-representation.

--
cody

[Freeware, Games and Humor]
www.deutronium.de.vu || www.deutronium.tk
Nov 13 '05 #27

P: n/a
pete wrote:
CBFalconer wrote:
pete wrote:
Nils Petter Vaskinn wrote:
> On Wed, 15 Oct 2003 01:59:04 -0700, RB wrote:
>
> > How to extract bytes from long, starting from the last byte?

#include <limits.h>
> #include <stdio.h>
>
> int main() {
>
> unsigned long value = 0x12345678;
> int i;
>
> printf("%#lx\n",value);
>
> for (i = sizeof value; i > 0; --i) {
> printf("%#x\n", value & 0xff);
> value >>= 8;

/*
** You realise that you don't know the size of value,
** so you might as well go all the way.
*/
printf("%#x\n", value & (unsigned char)-1);
value >>= CHAR_BIT;

> }
> return 0;
> }


You need neither CHAR_BIT nor shifts nor limits.h nor sizeof:

for (i = 8; i > 0; --i) {
printf("%x ", value % 256);
value /= 256;
}
putchar('\n'); /* <--AND HERE is where the \n goes */
return 0;
}

and the result is portable.


I was addressing the more general subject,
in the subject line of this thread: "How to extract bytes from long?",
rather than how to extract bytes from 0x12345678
or any other number which doesn't require more than 32 bits.


You were not extracting bytes. You were extracting 8 bit
quantities, least significant part first. In other words you are
expressing the _value_ in base 256, so why not say so in the code?

If the compiler knows that it can improve the code by using
shifts, it may do so.

--
Chuck F (cb********@yahoo.com) (cb********@worldnet.att.net)
Available for consulting/temporary embedded and systems.
<http://cbfalconer.home.att.net> USE worldnet address!
Nov 13 '05 #28

P: n/a

On Fri, 17 Oct 2003, Jirka Klaue wrote:

Arthur J. O'Dwyer wrote:
...
(uchar)-1 == (uchar)-((int)1) == (-(int)1)+UCHAR_MAX == UCHAR_MAX-1


You must have miscalculated somewhere. (uc)-1 should be UCHAR_MAX.


Augh! I do that every time! Thanks.

(uc)-1 == (uc)(0 - 1) == (-1+UCHAR_MAX+1) == UCHAR_MAX

-Arthur
Nov 13 '05 #29

P: n/a

On Fri, 17 Oct 2003, cody wrote:

"Arthur J. O'Dwyer" <aj*@nospam.andrew.cmu.edu> schrieb...
However, it *does* produce the right answer, which is a point in
its favor. ~0 might trap, and in any case I think (unsigned char)-1
has a bit more aesthetic value to it (YMMV, of course).
Why should ~0 trap??? it results in the 1's complement of 0 which
means all bits are 1's


....which may be a trap representation on a ones'-complement
architecture.
Padding bits are not affected by this operation, however the
values of padding bits should never be of your interest.
Well, technically padding bits *might* be affected by the ~
operation, but the effect on the padding bits alone cannot create
a trap representation -- the system has to remember to do the
Right Thing with them in this case.
A trap representation can *only* be generated when manipulating the
value using pointers which aren't the type of the value or doesn't start
at the exact address of the value.
Wrong. Signed integer overflow may create a trap value, for instance.
Wrong usage of an union can also result in a trap representation.

But all arithmetic or bitwise operations cannot result in a
trap representation.


Wrong.

-Arthur

Nov 13 '05 #30

P: n/a
On Fri, 17 Oct 2003 20:06:45 +0200, cody wrote:
"Arthur J. O'Dwyer" <aj*@nospam.andrew.cmu.edu> schrieb im Newsbeitrag
news:Pi**********************************@unix48.a ndrew.cmu.edu...
However, it *does* produce the right answer, which is a point in
its favor. ~0 might trap, and in any case I think (unsigned char)-1
has a bit more aesthetic value to it (YMMV, of course).


Why should ~0 trap??? it results in the 1's complement of 0 which means all
bits are 1's. Padding bits are not affected by this operation, however the
values of padding bits should never be of your interest.

A Trap representation can *only* be generated when manipulating the value
using pointers which aren't the type of the value or doesn't start at the
exact address of the value. Wrong usage of an union can also result in a
Trap representation.

But all arithmetic or bitwise operations cannot result in a
trap-representation.


Go back and read the whole thread

Nov 13 '05 #31

P: n/a
"pete" <pf*****@mindspring.com> wrote in message
news:3F***********@mindspring.com...
Peter Nilsson wrote:

....
UB if sizeof(long) == 1.


No.
If sizeof(long) == 1,
then the loop doesn't execute.


My bad.

--
Peter
Nov 13 '05 #32

P: n/a
CBFalconer wrote:

pete wrote:
CBFalconer wrote:
pete wrote:
> Nils Petter Vaskinn wrote:
> > On Wed, 15 Oct 2003 01:59:04 -0700, RB wrote:
> >
> > > How to extract bytes from long, starting from the last byte?
>
> #include <limits.h>
> > #include <stdio.h>
> >
> > int main() {
> >
> > unsigned long value = 0x12345678;
> > int i;
> >
> > printf("%#lx\n",value);
> >
> > for (i = sizeof value; i > 0; --i) {
> > printf("%#x\n", value & 0xff);
> > value >>= 8;
>
> /*
> ** You realise that you don't know the size of value,
> ** so you might as well go all the way.
> */
> printf("%#x\n", value & (unsigned char)-1);
> value >>= CHAR_BIT;
>
> > }
> > return 0;
> > }

You need neither CHAR_BIT nor shifts nor limits.h nor sizeof:

for (i = 8; i > 0; --i) {
printf("%x ", value % 256);
value /= 256;
}
putchar('\n'); /* <--AND HERE is where the \n goes */
return 0;
}

and the result is portable.


I was addressing the more general subject,
in the subject line of this thread:
"How to extract bytes from long?",
rather than how to extract bytes from 0x12345678
or any other number which doesn't require more than 32 bits.


You were not extracting bytes. You were extracting 8 bit
quantities, least significant part first. In other words you are
expressing the _value_ in base 256, so why not say so in the code?


I don't know what you're talking about.
Which code do you think is mine?

--
pete
Nov 13 '05 #33

P: n/a
Chris Torek <no****@elf.eng.bsdi.com> wrote in message news:<bm**********@elf.eng.bsdi.com>...
The section quoted above (along with others) define how the abtract
machine is to work. In the abstract machine, writing:

~0

means:

- make an int with the value 0
- now, flip all the bits

This process *can* give rise to a "trap representation" on a ones'
complement machine.


Thanks, I get it now. The correct usage is therefore ~0u or ~0U.

Sam
Nov 13 '05 #34

P: n/a
op*****@yahoo.com (Samuel Barber) wrote in message news:<37**************************@posting.google. com>...
pete <pf*****@mindspring.com> wrote in message news:<3F***********@mindspring.com>...
Samuel Barber wrote:

ai***@acay.com.au (Peter Nilsson) wrote in message news:<63**************************@posting.google. com>...
> op*****@yahoo.com (Samuel Barber) wrote in message news:<37**************************@posting.google. com>...
> > pete <pf*****@mindspring.com> wrote in message news:<3F***********@mindspring.com>...
> > > printf("%#x\n", value & (unsigned char)-1);
> > > for (i = sizeof value - 1; i != 0; --i) {
> > > value >>= CHAR_BIT;
> > > printf("%#x\n", value & (unsigned char)-1);
> > > }
> > > return 0;
> > > }
> >
> > I would use ~0 for "all 1s" rather than -1.
>
> That risks a trap representation under C99.

Nonsense. ~0 is idiomatic C. Aren't you worried about the "risk" that
-1 may not be implemented as all-1s? That is, after all, an
implementation detail. It's not true for sign-magnitude or 1's
complement, for example.


You're wrong about everything.

~0 is negative zero in ones complement.
Implementations are allowed trap negative zero.


You're just repeating words, without any understanding. There's no
such thing as "negative zero" (or negative anything) in the context of
bitwise operations. How can a bitwise operation trap? It can't.


I was wrong about everything. ~0 is "wrong" (in terms of the abstract
C machine); the correct expression is ~0u or ~0U.

Sam
Nov 13 '05 #35

P: n/a
"cody" <do*********************@gmx.de> writes:
"Arthur J. O'Dwyer" <aj*@nospam.andrew.cmu.edu> schrieb im Newsbeitrag
news:Pi**********************************@unix48.a ndrew.cmu.edu...
However, it *does* produce the right answer, which is a point in
its favor. ~0 might trap, and in any case I think (unsigned char)-1
has a bit more aesthetic value to it (YMMV, of course).


Why should ~0 trap??? it results in the 1's complement of 0 which means all
bits are 1's. Padding bits are not affected by this operation, however the
values of padding bits should never be of your interest.

A Trap representation can *only* be generated when manipulating the value
using pointers which aren't the type of the value or doesn't start at the
exact address of the value. Wrong usage of an union can also result in a
Trap representation.

But all arithmetic or bitwise operations cannot result in a
trap-representation.


This is not true. You should read the relevant portions of the
standard before making such assertions.

-Micah
Nov 13 '05 #36

P: n/a
"Arthur J. O'Dwyer" <aj*@nospam.andrew.cmu.edu> writes:
On Fri, 17 Oct 2003, cody wrote:

"Arthur J. O'Dwyer" <aj*@nospam.andrew.cmu.edu> schrieb...
However, it *does* produce the right answer, which is a point in
its favor. ~0 might trap, and in any case I think (unsigned char)-1
has a bit more aesthetic value to it (YMMV, of course).


Why should ~0 trap??? it results in the 1's complement of 0 which
means all bits are 1's


...which may be a trap representation on a ones'-complement
architecture.
Padding bits are not affected by this operation, however the
values of padding bits should never be of your interest.


Well, technically padding bits *might* be affected by the ~
operation, but the effect on the padding bits alone cannot create
a trap representation -- the system has to remember to do the
Right Thing with them in this case.


I don't believe this is true. Can you back this up with a quote
from the standard? I cannot remember any instance where the
standard says this can't be: and I can think of a specific spot
where the standard says (non-normatively) that this *can* be (see
footnote 45 to 6.2.6.2#5). The lack of exclusion means it
certainly is possible.

-Micah
Nov 13 '05 #37

P: n/a

On Sat, 18 Oct 2003, Micah Cowan wrote:

"Arthur J. O'Dwyer" writes:
On Fri, 17 Oct 2003, cody wrote:

Why should ~0 trap??? it results in the 1's complement of 0 which
means all bits are 1's
...which may be a trap representation on a ones'-complement
architecture.
Padding bits are not affected by this operation, however the
values of padding bits should never be of your interest.


Well, technically padding bits *might* be affected by the ~
operation, but the effect on the padding bits alone cannot create
a trap representation -- the system has to remember to do the
Right Thing with them in this case.


I don't believe this is true. Can you back this up with a quote
from the standard?


From N869, section 6.2.6.2, footnote 39:

Some combinations of padding bits might generate trap
representations, for example, if one padding bit is a
parity bit. Regardless, no arithmetic operation on valid
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
values can generate a trap representation other than as
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
part of an exception such as an overflow, and this cannot
occur with unsigned types. All other combinations of
padding bits are alternative object representations of
the value specified by the value bits.

(The index to N869 clearly puts "bitwise operators" under "arithmetic
operators," even if the text doesn't explicitly say so.)

This is obviously the Right Thing for the standard to say, too,
since if padding bits' values *could* create trap representations
out of thin air, how could programmers on those platforms ever
compute anything?
I cannot remember any instance where the
standard says this can't be: and I can think of a specific spot
where the standard says (non-normatively) that this *can* be (see
footnote 45 to 6.2.6.2#5).


N869 doesn't have a 6.2.6.2#5. Could you post that paragraph and
footnote, please?

-Arthur

Nov 13 '05 #38

P: n/a
"Arthur J. O'Dwyer" <aj*@nospam.andrew.cmu.edu> writes:
On Sat, 18 Oct 2003, Micah Cowan wrote:

"Arthur J. O'Dwyer" writes:
On Fri, 17 Oct 2003, cody wrote:
>
> Why should ~0 trap??? it results in the 1's complement of 0 which
> means all bits are 1's

...which may be a trap representation on a ones'-complement
architecture.

> Padding bits are not affected by this operation, however the
> values of padding bits should never be of your interest.

Well, technically padding bits *might* be affected by the ~
operation, but the effect on the padding bits alone cannot create
a trap representation -- the system has to remember to do the
Right Thing with them in this case.
I don't believe this is true. Can you back this up with a quote
from the standard?


From N869, section 6.2.6.2, footnote 39:

Some combinations of padding bits might generate trap
representations, for example, if one padding bit is a
parity bit. Regardless, no arithmetic operation on valid
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
values can generate a trap representation other than as
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
part of an exception such as an overflow, and this cannot
occur with unsigned types. All other combinations of
padding bits are alternative object representations of
the value specified by the value bits.

(The index to N869 clearly puts "bitwise operators" under "arithmetic
operators," even if the text doesn't explicitly say so.)


The above quote is footnote 45 (*and* 44: typo!) of
6.2.6.2 in the final version.

The same grouping appears in the index of the final version; but
the index isn't normative (nor are the footnotes). However, the
footnote you quote above still seems to be borne out by normative
text. But I don't think the quote quite supports your assertion
that "the padding bits alone cannot create a trap
representation": The fact that "proper" arithmetic operations
cannot affect the padding bits in such a way as to create a trap
representation doesn't imply that affecting the padding bits can
not be the sole cause of trap representations; it means that
implementations can't cause padding bits to somehow find
themselves in that representation in the coruse of well-defined
arithmetic operations. But this is obvious, because the only way
to affect padding bits in the first place is to use arithmetic
operations in such a way as to cause exceptional conditions (such
as overflows). I see no reason why using ~ on a signed integer
type can't cause such an exceptional condition: Obviously,
overflow itself can easily occur, and the very act of flipping a
padding bit could be considered an "exceptional condition",
AFAICT.
This is obviously the Right Thing for the standard to say, too,
since if padding bits' values *could* create trap representations
out of thin air, how could programmers on those platforms ever
compute anything?


This is not a valid conclusion. Assuming padding bits' values
*can* create trap representations out of thin air, you can only
conclude that an implementation can't arrange padding bits to
such a representation when arithmetic operators are being
*properly* used. Can you think of a well-defined way to directly
affect padding bits? But some ways that don't involve arithmetic
operators (directly) would be to read an uninitialized value
(which could already *be* a trap representation), or you could
do:

int foo;
unsigned char *bar = (unsigned char *)&foo; /* Well-defined */

bar[0] = bar[0] ^ 01000; /* Well-defined, but might set a
padding bit of foo. */
foo; /* Not well-defined: might be a trap representation. */
I cannot remember any instance where the
standard says this can't be: and I can think of a specific spot
where the standard says (non-normatively) that this *can* be (see
footnote 45 to 6.2.6.2#5).


N869 doesn't have a 6.2.6.2#5. Could you post that paragraph and
footnote, please?


The paragraph doesn't tell you anything you don't know (it's got
to be in N869), and the footnote is the very same one you've
quoted from N869. But here it is, anyway (6.2.6.2#5:)

The values of any padding bits are unspecified.45) A valid
(non-trap) object representation of a signed integer type where
the sign bit is zero is a valid object representation of the
corresponding unsigned type, and shall represent the same
value.

-Micah
Nov 13 '05 #39

P: n/a
Micah Cowan <mi***@cowan.name> wrote:
"Arthur J. O'Dwyer" <aj*@nospam.andrew.cmu.edu> writes:

<snip>
N869 doesn't have a 6.2.6.2#5. Could you post that paragraph and
footnote, please?


The paragraph doesn't tell you anything you don't know (it's got
to be in N869)

<snip>

Chapter 6.2.6.2:
C99 final N869
#1 --> #1
#2 --> #2
#3 --> --
#4 --> --
#5 --> #3
#6 --> #4

Regards
--
Irrwahn
(ir*******@freenet.de)
Nov 13 '05 #40

P: n/a
Samuel Barber wrote:
The correct usage is therefore ~0u or ~0U.


Bitwise operators are usually used on unsigned types.

Bitwise operators return implementation defined values
or cause undefined behavior, on signed types,
if the sign bit is set or if the operation attempts to set it.

--
pete
Nov 13 '05 #41

P: n/a
CBFalconer wrote:

Peter Shaggy Haywood wrote:
CBFalconer was jivin' on Thu, 16 Oct 2003 06:07:04 GMT
You need neither CHAR_BIT nor shifts nor limits.h nor sizeof:

for (i = 8; i > 0; --i) {
printf("%x ", value % 256);
value /= 256;
}
putchar('\n'); /* <--AND HERE is where the \n goes */
return 0;
}

and the result is portable.

Why not just:
puts("0x78, 0x56, 0x34, 0x12");
?
That's portable too.

Hardly. You are making huge assumptions about the sizes of long and
unsigned char. Who says sizeof(long) is 8? And who says a byte is 8
bits? These assumptions are evident in your code, and are decidedly
non-portable.
But it is, if you say the object is to emit 8 pairs of hex digits,


Where did you get the idea that the object,
is to emit 8 pairs of hex digits?
Nothing like that, has even been remotely suggested
anywhere previously in this thread.
with the first representing the least significant bits. This
gives a fixed output field. Conceded, it truncates anything past
64 bits in an unsigned long.

If the objective is to dump all digits, and ignore the more
significant zeroes, you can simply exit the loop when the value
becomes zero.

In both cases the technique is independant of CHAR_BIT and
sizeof(long) and, most important, independent of endianess.


The technique is also independant of, and irrelevant to,
the the subject line of this thread.
OP gave an example which suggested 8 bits per byte,
but the question was "How to extract bytes from long?"

--
pete
Nov 13 '05 #42

P: n/a
pete wrote:
CBFalconer wrote:
Peter Shaggy Haywood wrote:
CBFalconer was jivin' on Thu, 16 Oct 2003 06:07:04 GMT

> You need neither CHAR_BIT nor shifts nor limits.h nor sizeof:
>
> for (i = 8; i > 0; --i) {
> printf("%x ", value % 256);
> value /= 256;
> }
> putchar('\n'); /* <--AND HERE is where the \n goes */
> return 0;
> }
>
> and the result is portable. .... snip ...
Hardly. You are making huge assumptions about the sizes of long and
unsigned char. Who says sizeof(long) is 8? And who says a byte is 8
bits? These assumptions are evident in your code, and are decidedly
non-portable.
But it is, if you say the object is to emit 8 pairs of hex digits,


Where did you get the idea that the object,
is to emit 8 pairs of hex digits?
Nothing like that, has even been remotely suggested
anywhere previously in this thread.


From the original post, quoted below: How to extract bytes from long, starting from the last byte?
For example, I have a long number:
0x12345678
I need to represent it as the following bytes list:
0x78, 0x56, 0x34, 0x12


Granted, it is not definitive, but is how I interpreted it.

--
Chuck F (cb********@yahoo.com) (cb********@worldnet.att.net)
Available for consulting/temporary embedded and systems.
<http://cbfalconer.home.att.net> USE worldnet address!
Nov 13 '05 #43

P: n/a
CBFalconer wrote:

pete wrote:
CBFalconer wrote:
Peter Shaggy Haywood wrote:
> CBFalconer was jivin' on Thu, 16 Oct 2003 06:07:04 GMT
>
> > You need neither CHAR_BIT nor shifts nor limits.h nor sizeof:
> >
> > for (i = 8; i > 0; --i) {
> > printf("%x ", value % 256);
> > value /= 256;
> > }
> > putchar('\n'); /* <--AND HERE is where the \n goes */
> > return 0;
> > }
> >
> > and the result is portable.
.... snip ...
But it is, if you say the object is to emit 8 pairs of hex digits,


Where did you get the idea that the object,
is to emit 8 pairs of hex digits?
Nothing like that, has even been remotely suggested
anywhere previously in this thread.

From the original post, quoted below:

How to extract bytes from long, starting from the last byte?
For example, I have a long number:
0x12345678
I need to represent it as the following bytes list:
0x78, 0x56, 0x34, 0x12


Granted, it is not definitive, but is how I interpreted it.


That's only 4 pairs of hex digits.
This is your output:
78 56 34 12 0 0 0 0

--
pete
Nov 13 '05 #44

P: n/a
Samuel Barber wrote:

op*****@yahoo.com (Samuel Barber) wrote in message news:<37**************************@posting.google. com>...
pete <pf*****@mindspring.com> wrote in message news:<3F***********@mindspring.com>...
Samuel Barber wrote:
>
> ai***@acay.com.au (Peter Nilsson) wrote in message news:<63**************************@posting.google. com>...
> > op*****@yahoo.com (Samuel Barber) wrote in message news:<37**************************@posting.google. com>...
> > > pete <pf*****@mindspring.com> wrote in message news:<3F***********@mindspring.com>...
> > > > printf("%#x\n", value & (unsigned char)-1);
> > > > for (i = sizeof value - 1; i != 0; --i) {
> > > > value >>= CHAR_BIT;
> > > > printf("%#x\n", value & (unsigned char)-1);
> > > > }
> > > > return 0;
> > > > }
> > >
> > > I would use ~0 for "all 1s" rather than -1.
> >
> > That risks a trap representation under C99.
>
> Nonsense. ~0 is idiomatic C.
> Aren't you worried about the "risk" that
> -1 may not be implemented as all-1s? That is, after all, an
> implementation detail. It's not true for sign-magnitude or 1's
> complement, for example.

You're wrong about everything.

~0 is negative zero in ones complement.
Implementations are allowed trap negative zero.


You're just repeating words, without any understanding. There's no
such thing as "negative zero" (or negative anything)
in the context of
bitwise operations. How can a bitwise operation trap? It can't.


I was wrong about everything. ~0 is "wrong" (in terms of the abstract
C machine); the correct expression is ~0u or ~0U.


(unsigned char)~0u would work instead of
(unsigned char)-1, in the above code example,
but for the general case of casting a constant,
to get an object with all bits set,
(unsigned cast)-1 is more versatile than (unsigned cast)~0u.
((size_t)-1) and ((long unsigned)-1),
are both unsigned values with all bits set.
((size_t)~0u) and ((long unsigned)~0u),
may or may not have all bits set,
depending on whether the width of the respective types,
is greater than the width of unsigned.

--
pete
Nov 13 '05 #45

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