ARM NEON:比较128位值 [英] ARM NEON: comparing 128 bit values

问题描述

我有兴趣找到一种最快的方式(最低的周期计数)，以比较存储在Cortex-A9内核(允许VFP指令)上的NEON寄存器(例如Q0和Q3)中的值.

I'm interested in finding the fastest way (lowest cycle count) of comparing the values stored into NEON registers (say Q0 and Q3) on a Cortex-A9 core (VFP instructions allowed).

到目前为止，我有以下内容:

So far I have the following:

(1)使用VFP浮点比较:

(1) Using the VFP floating point comparison:

vcmp.f64        d0, d6
vmrs            APSR_nzcv, fpscr
vcmpeq.f64      d1, d7
vmrseq          APSR_nzcv, fpscr

如果64位浮点数"与NaN等效，则此版本将不起作用.

If the 64bit "floats" are equivalent to NaN, this version will not work.

(2)使用NEON缩小和VFP比较(这次仅一次且以NaN安全的方式进行):

(2) Using the NEON narrowing and the VFP comparison (this time only once and in a NaN-safe manner):

vceq.i32        q15, q0, q3
vmovn.i32       d31, q15
vshl.s16        d31, d31, #8
vcmp.f64        d31, d29
vmrs            APSR_nzcv, fpscr

D29寄存器先前已预先装入了正确的16位模式:

The D29 register is previously preloaded with the right 16bit pattern:

vmov.i16        d29, #65280     ; 0xff00

我的问题是:还有什么比这更好的了吗?我正在监督一些明显的方法吗?

My question is: is there any better than this? Am I overseeing some obvious way to do it?

推荐答案

我相信您可以通过一条指令来减少它.通过使用左移并插入(VLSI)，可以将Q15的4个32位值组合为D31中的4个16位值.然后，您可以将其与0进行比较并获得浮点标志.

I believe you can reduce it by one instruction. By using the shift left and insert (VLSI), you can combine the 4 32-bit values of Q15 into 4 16-bit values in D31. You can then compare with 0 and get the floating point flags.

vceq.i32  q15, q0, q3
vlsi.32   d31, d30, #16
vcmp.f64  d31, #0
vmrs      APSR_nzcv, fpscr

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