Hello Niels,
> > This is where after a lot of scratching of my thinking cap I got to the
> > conclusion that in LE we're actually working with the least significant
> > bytes of r4 at the low end of the register.
> My understanding of LE here is that the least significant CNT bits
> (first in memory) where processed earlier, and the remaining TNC bits we
> need to process are at the high end. So we first shift right CNT bits to
> discard the bits already processed, and then do eight bit a a time,
> shifting right in the loop.
Okay, so the comment is referring to the situation in the register
literally just before the next instruction. I was clearly looking too
far afield for clues on what's going on.
> > My guess is that it's just a matter of interpretation what end of the
> > register is "high" and most significant.
> The way I think about it, a 32-bit register holds a binary integer. Each
> bit has a weight, from 1 to 2^31 (let's stick to unsigned
> interpretation). And then I try to stay with the widely used convention
> that bits are numbered 0-31, with bit k having weight 2^k, and with
> "right", "lower", "less significant", being synonymous, all meaning the
> bits with smaller weight, and "left", "higher", "more significant"
> meaning bits with higher weight.
> Note this termonology is endian independent. It's only when storing the
> integer in memory on a byte-addressed machine, that it becomes relevant
> to ask which 8 bits get stored at which address, with "little-endian"
> meaning that the lower/right/less significant bits in the register get
> stored at lower addresses in memory.
> Maybe it's illuminating to compare with bit order. On a byte-addressed
> machine, we can't really talk about in which "order" individual bits of
> a byte are stored in memory. But we'd have to care about bit order,
> e.g., if transmitting a byte serially over a wire.
Thanks for indulging me. :) I had a similar understanding and treatise
to that effect all typed up but then found ARMs view of the world in
their ARM and felt stupid.
> > Long story short: Let me know what to do with those comments based on
> > how much my thinking is off.
> You could maybe just say "We have TNC/8 left-over bytes in r4 (high end
> if little endian, low end if big endian)". Or feel free to rephrase.
I've gone with:
diff --git a/arm/memxor.asm b/arm/memxor.asm
index 239a4034..e4619629 100644
--- a/arm/memxor.asm
+++ b/arm/memxor.asm
@@ -138,24 +138,25 @@ PROLOGUE(nettle_memxor)
adds N, #8
beq .Lmemxor_odd_done
- C We have TNC/8 left-over bytes in r4, high end
+ C We have TNC/8 left-over bytes in r4, high end on LE and low end on
+ C BE, excess bits to be discarded by alignment adjustment at the other
S0ADJ r4, CNT
+ C now byte-aligned at low end on LE and high end on BE
ldr r3, [DST]
eor r3, r4
Patch in followup mail for your consideration.
--
Best wishes,
Michael
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