>> Then again, if all the world were big-endian, the issue would be moot :)
> But even on big-endian architectures, registers tend to be little-endian
little-endian. To see this, consider the following
move two-byte integer from A to X
move one-byte integer from X to B
Question: will the byte at B end up containing the high byte or
the low byte of A? In little-endian architectures, the answer is
always "the low byte". However, in big-endian architectures, the
answer depends on whether X is a memory location or a register; if
it's a memory location, then B gets the high byte. Otherwise, it
gets the low byte.
You should have mentioned that /you/ wrote that text.
I'm pretty sure you're wrong in any normal sense of the word, and that
registers are endian-less ... but if you want to argue that you're
/not/, first show how it matters to the programmer. For example by
showing a function which detects if your CPU has big- or little-endian
 Not counting odd x86 registers where register FOO is really 16
bits of the 32-bit register BAR and so on ... that sort of thing
was clearly not what you had in mind.
\X/ snipabacken.se> O o .