有人可以解释这个 SSE BigNum 比较吗? [英] can someone explain this SSE BigNum comparison?

问题描述

如果你看看这个 answer，作者设法为存储在 2 个 SSE 寄存器中的 2 个整数 bignum 创建了一个紧凑的比较算法.我没有很好地遵循它:)

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 :)

到目前为止我做了什么:

What I did so far:

if l = a 和

e = a == b = {a[i] == b[i] ? ~0 : 0}

然后 a

但这似乎不是作者在做什么.我错过了什么?有什么需要大于比较的?

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?

推荐答案

我忽略了答案不是通用的，而是仅限于由 32 位元素组成的 64 位 bignum.如果您有 2 个 64 位向量 a = {a0, a1}, b = {b0, b1}，则程序计算:

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)

在我的问题中，我的目标是使用 SSE/AVX 寄存器实现的任意长的 BigNum.

In my question I was aiming at arbitrarily long BigNums, implemented with SSE/AVX registers.

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