Re: [9fans] xd

[erik quanstrom <quanstro@quanstro.net>]

On Wed May 20 06:57:14 EDT 2009, uair00@gmail.com wrote:
> I have an xd(1) question. Am I wrong or xd gets the byte ordering wrong?

no.  xd is correct.  if you're running on an intel,
you're running on a little-endian machine which
means that numbers are stored in the reverse order
they are written.

#include <u.h>
#include <libc.h>

void
main(void)
{
	uchar e[8] = {0, 1, 2, 3, 4, 5, 6, 7};
	int i;
	uvlong l;

	l = *(uvlong*)e;
	print("%.16llux\n", l);

	l = 0x01020304050607ull;
	memcpy(e, &l, 8);
	for(i = 0; i < nelem(e); i++)
		print("%.2ux", e[i]);
	print("\n");
}

see http://en.wikipedia.org/wiki/Endianness

> 2. xd output from p9p shows exactly the opposite byte ordering that
> hexdump output.
> Perhaps there's something wrong with xd.

neither is wrong.  hexdump is just underspecified.  hexdump
doesn't say what the endianness of its output is.  xd on the other
hand does:

          Formats other than -c are specified by pairs of characters
          telling size and style, `4x' by default.  The sizes are

          1 or b   1-byte units.
          2 or w   2-byte big-endian units.
          4 or l   4-byte big-endian units.
          8 or v   8-byte big-endian units.

so numbers will be printed in reverse on an intel machine.
but the same network packet will be printed the same way
by xd on a big-endian sender and a little-endian recipient.

- erik