手臂霓虹灯比较操作产生负一 [英] arm neon compare operations generate negative one

问题描述

我想下面的汇编code：

I am trying the following assembly code:

vclt.f32 q9,q0,#0
vst1.i32 q9,[r2:128]

但是，如果条件为真，在Q9相应的元素设置为负一，而不是积极的。

But if the condition is true, the corresponding element in q9 is set to negative one instead of positive one.

我能做些什么，得到了肯定的？

What can I do to get a positive one ?

推荐答案

有不在NEON很多有条件的东西，但什么也真的只有按位可行的，而不是布尔逻辑 - 例如见 VBSL 。

There's not a lot of conditional stuff in NEON, but what there is is really only workable with bitwise, rather than Boolean, logic - see e.g. vbsl.

如果你有基本的可怕回忆，真的很讨厌按位真值，那么琐碎的方式向蒙版转换为布尔是只取每个元素的最高位：

If you have horrible memories of BASIC and really hate bitwise truth values, then the trivial way to convert the mask to a Boolean is to just take the top bit of each element:

vshr.u32 q9, q9, #31

虽然否定，而可以说不太清楚，一目了然阅读，可能是微观更好的性能，明智的在某些情况下：

Although negation, whilst arguably less clear to read at a glance, could be microscopically better performance-wise in some cases:

vneg.s32 q9, q9

（从通过微架构的时间的浏览，这两个操作都pretty很多是相同的，但 VNEG 的一些理论优于 vshr 是，它消耗其输入以后的Cortex-A8，并且能够发出下来两个ASIMD管道的Cortex-A57 / A72）

(from a browse through microarchitectural timings, both operations are pretty much identical, but some theoretical advantages of vneg over vshr are that it consumes its inputs later on Cortex-A8, and can issue down both ASIMD pipes of Cortex-A57/A72)

无论哪种方式，如在顶部说，这只有真正有意义存储结果存储在存储器非矢量化code来看待。

Either way, as said at the top, this only really makes sense for storing the result back to memory to be looked at by non-vectorised code.

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