问题
If you look at this answer, the author manages to create a compact comparison algorithm for 2 integer bignums, stored in 2 SSE registers. I am not following it too well :)
What I did so far:
if l = a < b = {a[i] < b[i] ? ~0 : 0} and
e = a == b = {a[i] == b[i] ? ~0 : 0}
then a < b == l[3] v e[3]l[2] v e[3]e[2]l[1] v e[3]e[2]e[1]l[0]
But this does not seem to be what the author is doing. What am I missing? What need is there for a greater than comparison?
回答1:
I've overlooked than the answer was not generic, but limited to 64-bit bignums, composed out of 32-bit elements. If you have 2 64-bit vectors a = {a0, a1}, b = {b0, b1}, then the program calculates:
a < b = ((a1 < b1) | (a0 < b0)) & ~(a1 > b1)
In my question I was aiming at arbitrarily long BigNums, implemented with SSE/AVX registers.
来源：https://stackoverflow.com/questions/27929402/can-someone-explain-this-sse-bignum-comparison