如何从__m64值的lsb创建8位掩码? [英] How to create a 8 bit mask from lsb of __m64 value?

问题描述

我有一个用例，其中我有一个位数组，例如，每个位都表示为8位整数.uint8_t data[] = {0,1,0,1,0,1,0,1};我想通过仅提取每个值的lsb来创建一个整数.我知道使用int _mm_movemask_pi8 (__m64 a)函数可以创建掩码，但是此内在函数仅占用字节的msb而不是lsb.是否有类似的内在方法或有效方法来提取lsb以创建单个8位整数?

I have a use case, where I have array of bits each bit is represented as 8 bit integer for example uint8_t data[] = {0,1,0,1,0,1,0,1}; I want to create a single integer by extracting only lsb of each value. I know that using int _mm_movemask_pi8 (__m64 a) function I can create a mask but this intrinsic only takes a msb of a byte not lsb. Is there a similar intrinsic or efficient method to extract lsb to create single 8 bit integer?

推荐答案

没有直接的方法，但是显然您可以简单地将lsb移入msb，然后将其提取:

There is no direct way to do it, but obviously you can simply shift the lsb into the msb and then extract it:

_mm_movemask_pi8(_mm_slli_si64(x, 7))

这几天使用MMX很奇怪，应该避免使用.

Using MMX these days is strange and should probably be avoided.

这是SSE2版本，仍仅读取8个字节:

Here is an SSE2 version, still reading only 8 bytes:

int lsb_mask8(uint8_t* bits) {
    __m128i x = _mm_loadl_epi64((__m128i*)bits);
    return _mm_movemask_epi8(_mm_slli_epi64(x, 7));
}

使用SSE2代替MMX避免了EMMS

Using SSE2 instead of MMX avoids the needs for EMMS

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