On Aug 25, Ken Nakata wrote
> Ahh...  This is what we differ from each other.
> 
> IDE bus is a little-endian 16-bit wide bus, so the string "Ne"
> (assuming 16-bit aligned) is represented by a bit pattern 0x654e (byte
> 0 "N" is in the lower 8 bits, and byte 1 "e" in the upper 8 bits).
> Now, the same string "Ne" on the big-endian host bus is represented by
> a different bit pattern 0x4e65 (byte 0 "N" is in the upper 8 bits and
> byte 1 "e" in the lower 8 bits).  Therefore, string's byte order _is_
> preserved when bytes are swapped.
> 
> Make sense to you?

No. A string is endianless, it's only an array of chars.
Maybe an example is needed.

Maybe an example will help.
Consider the following C program:

main()
{
	char *s = "abc";
	int *i = (int*)s;

	printf("char values for %s: %x %x %x %x\n", s, 
	    s[0], s[1], s[2], s[3]);
	printf("int value: %x\n", *i);
}

That is, we have a memory location of 4 bytes, where is stored the string
"abc\0"

This memory location is interpreted in 2 ways: as an array of chars and
as an integer by the C program.
On my i386 I get:
manu:/tmp>./a.out 
./a.out
char values for abc: 61 62 63 0
int value: 636261

On my sparc (same endian as m68k) I get:
cordouan:/tmp>./a.out 
char values for abc: 61 62 63 0
int value: 61626300

So the same string have its bytes stored in the same order on both
CPUs. But these 4 bytes gives a different integer.
So if I want to transfer a string between these 2 machines I need to copy
the bytes in the same order, if I want to tranfer a int I need to copy
the bytes in reversed order.

--
Manuel Bouyer, LIP6, Universite Paris VI.           Manuel.Bouyer@lip6.fr
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