Re: Pointing to high and low bytes of something
From: mike gillmore (rmgillmore_at_hotmail.com)
Date: 11/20/03
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Date: Thu, 20 Nov 2003 03:52:15 GMT
"Chris Torek" <nospam@elf.eng.bsdi.com> wrote in message
news:bph50o$eh0$1@elf.eng.bsdi.com...
> In article <3FBBBEF0.87B2475F@sun.com>
> Eric Sosman <Eric.Sosman@Sun.COM> writes:
> >... I'm saying that (1) bit order is not detectable by any C
> >construct I can imagine, (2) bit order is not detectable by any CPU
> >instruction on a machine that lacks bit-level addressing,
> >(3) by Occam's Razor, that which is undetectable is better
> >omitted from discussion.
>
> Indeed. One can, however, bring up C's bitfield-in-structure
> construct:
>
> struct bits { int a:1, b:1, c:1 /* fill in more */ ; };
>
> which might *seem* to expose the hardware's bit order.
>
> In fact, it does not -- the bitfields are allocated by the
> compiler, and two different C compilers on the same hardware
> will sometimes use different bit orders.
>
> Even for case (2), sometimes the CPUs themselves exhibit a split
> personality. The Motorola 680x0 series did this: the single-bit
> instructions that operate on D registers (e.g., BIT #3, D0) use
> the opposite order from the bitfield instructions (e.g., BFEXT).
>
> [modern modem example, snipped]
> > The question: What is the "bit order" of the N bits
> >encoded by one single BZZZ-to-BEEP transition in this scheme?
> >Note that all N bits leave the transmitter encoded in one
> >single event and arrive at the receiver the same way: they
> >are simultaneous and indivisible -- and I say the entire
> >idea of "bit order" in such a situation is meaningless.
>
> Correct. Sequencing cannot arise unless there is a sequence. If
> there is no defined time or space division -- if all is an
> indistinguishable, atomic lump -- then the notion of "the part on
> the left" or "the part at the front of the queue" makes no sense.
>
> Of course, in C code, we (programmers) can take apart any byte
> (unsigned char) value in any way we like, using shifts and masks:
> val & 0x80, val & 0x01, val & 0x04, val & 0x02, val & 0x10, ...
> defines a bit order. But *we* have defined it, and thus we control
> the horizontal and the vertical. Only if you allow someone else
> to define the order -- say, by taking a two-byte object (one with
> sizeof obj == 2) and addressing it as two separate "unsigned char"s
> -- have you given up control; only then do you need to beg the one
> to whom you gave up that control: "pretty please, tell me the order
> *you* used, so that I may accomodate you". As Humpty Dumpty put
> it, the question is who is to be the master. :-)
> --
> 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://67.40.109.61/torek/index.html (for the
moment)
> Reading email is like searching for food in the garbage, thanks to
spammers.
I have used this little program for many years to discover the machine
endian-ness. Use it in good health.
#include <stdio.h>
int
main ( )
{
/*
* This short program will simply determine if this machine is
* a little endian or a big endian byte ordering architecture.
*
* If this machine is little endian, this program returns a zero. If
* this is big endian machine, this program returns non-zero
*/
unsigned short testValue = 0xdead;
unsigned char * firstBytePtr = ( unsigned char * )&testValue;
int isBigEndian;
isBigEndian = ( *firstBytePtr != ( unsigned char )testValue );
printf( " %x %s %x isBigEndian = %s(%d)\n",
*firstBytePtr, isBigEndian ? "!=" : "==", ( unsigned char )testValue,
isBigEndian ? "TRUE" : "FALSE", isBigEndian );
return ( isBigEndian );
} /* main() */
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