_mm_packus_epi16

最新推荐文章于 2023-11-20 21:37:41 发布

YiYueHuan

最新推荐文章于 2023-11-20 21:37:41 发布

阅读量3.2k

点赞数 1

CC 4.0 BY-SA版权

分类专栏： C/C++ SIMD C

本文链接：https://blog.youkuaiyun.com/yiyuehuan/article/details/50204819

C/C++ 同时被 3 个专栏收录

19 篇文章

订阅专栏

12 篇文章

订阅专栏

SIMD

1 篇文章

订阅专栏

本文详细解释了SSE指令集中的_mm_packus_epi16函数，包括其如何将16位整数饱和压缩到8位，并提供了示例代码来展示如何打印__m128i变量。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Microsoft Specific

Packs the 16 signed 16-bit integers from a and b into 8-bit unsigned integers and saturates.

 
__m128i _mm_packus_epi16 (__m128i a, __m128i b);
PACKUSWB

Return Value

 
      r0 := UnsignedSaturate(a0)
r1 := UnsignedSaturate(a1)
...
r7 := UnsignedSaturate(a7)
r8 := UnsignedSaturate(b0)
r9 := UnsignedSaturate(b1)
...
r15 := UnsignedSaturate(b7)

Requirements

Header: emmintrin.h

END Microsoft Specific

https://msdn.microsoft.com/library/07ad1wx4(v=vs.110).aspx

---------------------------------------------------------------------------------

What does UnsignedSaturate in SSE instruction mean?

Basically, "saturation" means that values beyond some "max" value get set to "max", and values below a "min" value get set to "min". Usually, "min" and "max" are the values appropiate for some data type.

Thus, for example, if you take arithmetic on unsigned bytes, "128+128" would have to be "256" (which is hex 0x100), which doesn't fit into a byte. Normal integer arithmetic would create an overflow and discard the part that doesn't fit, which means "128+128 -> 0". With saturated arithmetic, "256 > 255" so the result is 255.

Another option would be scaling, which basically "compresses" the values to a smaller range. Saturation just cuts them off.

You can also use this to put larger types into smaller ones, like putting 16 bit values into 8 bit values. Your example most likely does exactly that, although you'll probably know better than I do what kind of types you are dealing with there.

"UnsignedSaturation" most likely has a min of "0" and a "max" of whatever the max of the result type is. Thus, negative inputs get turned into "0".

http://stackoverflow.com/questions/12141075/what-does-unsignedsaturate-in-sse-instruction-mean

------------------------------------------------------------------------------------------------

how to print a __m128i variable?

#include <iostream>
#include <sstream>
#include <emmintrin.h>
#include <stdint.h>

using namespace std;

template <typename T>
string __m128i_toString(const __m128i var)
{
    stringstream sstr;
    const T *values = (const T *)&var;
    if (sizeof(T) == 1)
    {
        for (unsigned int i = 0; i < sizeof(__m128i); i++)
        {
            sstr << (int)values[i] << " ";
        }
    }
    else
    {
        for (unsigned int i = 0; i < sizeof(__m128i) / sizeof(T); i++)
        {
            sstr << values[i] << " ";
        }
    }
    return sstr.str();
}

int main()
{
    __m128i resultLo = _mm_setr_epi16(800, 700, 600, 500, 400, 300, 200, 100);
    __m128i resultHi = _mm_setr_epi16(0, -100, -200, -300, -400, -500, -600, -700);
    __m128i result = _mm_packus_epi16(resultLo, resultHi);

    cout << __m128i_toString<int16_t>(resultLo) << endl;
    cout << __m128i_toString<int16_t>(resultHi) << endl;
    cout << __m128i_toString<uint8_t>(result) << endl;

    return 0;
}

http://stackoverflow.com/questions/13257166/print-a-m128i-variable