[email protected] wrote:
/* Majority: (x^y)|(y&z)|(z&x) = (x & z) + ((x ^ z) & y)
#define F3(x,y,z,dest) \
movl z, TMP; \
andl x, TMP; \
addl TMP, dest; \
movl z, TMP; \
xorl x, TMP; \
andl y, TMP; \
addl TMP, dest
Since y is the most recently computed result (it's rotated in the
previous round), I arranged the code to delay its use as late as
possible.
Now you have one more register to play with.
Okay, thanks. It doesn't actually give one more register except in the
F3 rounds (TMP2 is normally used to hold the magic constants) but it's a
good cleanup.
A faster way is to unroll 5 iterations and do:
e += F(b, c, d) + K + rol32(a, 5) + W[i ]; b = rol32(b, 30);
d += F(a, b, c) + K + rol32(e, 5) + W[i+1]; a = rol32(a, 30);
c += F(e, a, b) + K + rol32(d, 5) + W[i+2]; e = rol32(e, 30);
b += F(d, e, a) + K + rol32(c, 5) + W[i+3]; d = rol32(d, 30);
a += F(c, d, e) + K + rol32(b, 5) + W[i+4]; c = rol32(c, 30);
then loop over that 4 times each. This is somewhat larger, but
still reasonably compact; only 20 of the 80 rounds are written out
long-hand.
I got this code from Nettle, originally, and I never looked at the SHA-1
round structure very closely. I'll give that approach a try.
Thanks
--Benjamin Gilbert
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