将8个16位SSE寄存器转换为8位数据 [英] Convert 8 16 bit SSE register to 8bit data

问题描述

考虑到我有一个16位数据的SSE阵列:

Consider I have an SSE array with 16 bit data:

{1,2,3,4,5,6,7,8}

现在，我需要通过在前8个字节中仅存储16位数据的低字节来将此SSE数组转换为8位数据，例如:

Now I need to convert this SSE array into 8 bit data by storing only the lower byte of 16 bit data in the first 8 bytes like:

{1,2,3,4,5,6,7,8,0,0,0,0,0,0,0,0}.

是否有任何SSE指令来执行此操作?

Is there any SSE instruction to perform this operation?

推荐答案

如 @harold 所述，您可以使用 pshufb aka轻松完成此操作_mm_shuffle_epi8 ，例如

As @harold says in the comments above, you can do this quite easily with pshufb aka _mm_shuffle_epi8, e.g.

#include <stdio.h>
#include <tmmintrin.h>

static __m128i pack_16_to_8(const __m128i v)
{
    const __m128i vperm = _mm_setr_epi8(0, 2, 4, 6, 8, 10, 12, 14, -1, -1, -1, -1, -1, -1, -1, -1);

    return _mm_shuffle_epi8(v, vperm);
}

int main(void)
{
    const __m128i v = _mm_setr_epi16(1, 2, 3, 4, 5, 6, 7, 8);

    printf("%vhd -> %vd\n", v, pack_16_to_8(v));
    return 0;
}

编译并运行:

$ gcc -Wall -mssse3 pack_16_to_8.c && ./a.out

1 2 3 4 5 6 7 8 -> 1 2 3 4 5 6 7 8 0 0 0 0 0 0 0 0

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