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社区首页 >问答首页 >英特尔本质:相乘交错8位值

问英特尔本质:相乘交错8位值
EN

Stack Overflow用户
提问于 2014-09-29 09:09:47
回答 2查看 1.9K关注 0票数 3

我正在处理一个RGBA32缓冲区(每个组件8位)，我需要将每个组件乘以一个常量，然后将乘法的每一个结果以这样的方式添加到其他部分：

结果= r*x +g*y+b*z+ a*w (两个向量之间的点积rgba和xyzw)

我试图使用英特尔SSE的本质来加速这个过程，但是我不知道如何在不改变输入的情况下做这样的事情。

有办法这样做吗？就像建立一个包含{x，y，z，w，x，y，z，w}的寄存器，并执行8位的饱和乘法？

最后的目标是将RGBA矢量乘以相应的颜色转换矩阵：

代码语言：javascript
复制
[ 66 129  25 0]   [R]
[-38 -74 112 0] * [G]
[112 -94 -18 0]   [B]
[0     0   0 0]   [A]

谢谢

编辑1:这是最后一个函数，使用浮点计算来获得更高的颜色精度，它使用SSE将rgba图像转换为YUV444图像。函数需要1.9到3.5ms之间的时间才能在intel i5 3570k上转换一个完整的高清映像，只需使用一个线程(很容易执行此功能，并且可以产生显著的性能改进)：

代码语言：javascript
复制
void SSE_rgba2YUV444_FP(char* a, char* y, char* u, char* v)
{
    __m128i mask = _mm_setr_epi8(0x00,0x04,0x08,0x0c, 0x01,0x05,0x09,0x0d, 0x02,0x06,0x0a,0x0e, 0x03,0x07,0x0b,0x0f); // Masque de mélange, chaque uint8 donne la position à donner (en offset en octet) du uint8 correspondant
    float m[9] = {0.299, 0.587, 0.114, -0.1687, -0.3313, 0.5, 0.5, -0.4187, -0.0813};                                                         // Dans le __m128i que l'on mélange

    __m128i row[4];
    for(int i=0; i<4; i++) {
        row[i] = _mm_loadu_si128((__m128i*)&a[16*i]);
        row[i] = _mm_shuffle_epi8(row[i],mask);
    }
    // row[i] = {rrrrggggbbbbaaaa} tous en uint8t
    __m128i t0 = _mm_unpacklo_epi32(row[0], row[1]); //to = {rrrrrrrrgggggggg}
    __m128i t1 = _mm_unpacklo_epi32(row[2], row[3]); //t1 = {rrrrrrrrgggggggg}
    __m128i t2 = _mm_unpackhi_epi32(row[0], row[1]); //t2 = {bbbbbbbbaaaaaaaa}
    __m128i t3 = _mm_unpackhi_epi32(row[2], row[3]); //t3 = {bbbbbbbbaaaaaaaa}
    row[0] = _mm_unpacklo_epi64(t0, t1); // row[0] = {rrrrrrrrrrrrrrrr}
    row[1] = _mm_unpackhi_epi64(t0, t1); // etc
    row[2] = _mm_unpacklo_epi64(t2, t3);

    __m128i v_lo[3], v_hi[3];
    for(int i=0; i<3; i++) {
        v_lo[i] = _mm_unpacklo_epi8(row[i],_mm_setzero_si128()); // On entrelace chaque row avec des 0, ce qui fait passer les valeurs
        v_hi[i] = _mm_unpackhi_epi8(row[i],_mm_setzero_si128()); // de 8bits à 16bits pour pouvoir travailler dessus
    }

    __m128 v32_lo1[3], v32_hi1[3], v32_lo2[3], v32_hi2[3];
    for(int i=0; i<3; i++) {
        v32_lo1[i] = _mm_cvtepi32_ps(_mm_unpacklo_epi16(v_lo[i],_mm_setzero_si128()));
        v32_lo2[i] = _mm_cvtepi32_ps(_mm_unpackhi_epi16(v_lo[i],_mm_setzero_si128()));
        v32_hi1[i] = _mm_cvtepi32_ps(_mm_unpacklo_epi16(v_hi[i],_mm_setzero_si128()));
        v32_hi2[i] = _mm_cvtepi32_ps(_mm_unpackhi_epi16(v_hi[i],_mm_setzero_si128()));
    } // On a nos rgb sur 32 bits

    __m128i yuv[3]; // {Y, U, V} 
    __m128 ylo1 = _mm_add_ps(_mm_mul_ps(v32_lo1[0], _mm_set1_ps(m[0])), _mm_add_ps(_mm_mul_ps(v32_lo1[1], _mm_set1_ps(m[1])), _mm_mul_ps(v32_lo1[2], _mm_set1_ps(m[2]))));
    __m128 ylo2 = _mm_add_ps(_mm_mul_ps(v32_lo2[0], _mm_set1_ps(m[0])), _mm_add_ps(_mm_mul_ps(v32_lo2[1], _mm_set1_ps(m[1])), _mm_mul_ps(v32_lo2[2], _mm_set1_ps(m[2]))));
    __m128 yhi1 = _mm_add_ps(_mm_mul_ps(v32_hi1[0], _mm_set1_ps(m[0])), _mm_add_ps(_mm_mul_ps(v32_hi1[1], _mm_set1_ps(m[1])), _mm_mul_ps(v32_hi1[2], _mm_set1_ps(m[2]))));
    __m128 yhi2 = _mm_add_ps(_mm_mul_ps(v32_hi2[0], _mm_set1_ps(m[0])), _mm_add_ps(_mm_mul_ps(v32_hi2[1], _mm_set1_ps(m[1])), _mm_mul_ps(v32_hi2[2], _mm_set1_ps(m[2]))));

    __m128i ylo1i = _mm_cvtps_epi32(ylo1);
    __m128i ylo2i = _mm_cvtps_epi32(ylo2);
    __m128i yhi1i = _mm_cvtps_epi32(yhi1);
    __m128i yhi2i = _mm_cvtps_epi32(yhi2);

    __m128i ylo = _mm_packus_epi32(ylo1i, ylo2i);
    __m128i yhi = _mm_packus_epi32(yhi1i, yhi2i);

    yuv[0] = _mm_packus_epi16(ylo, yhi);

    ylo1 = _mm_add_ps(_mm_add_ps(_mm_mul_ps(v32_lo1[0], _mm_set1_ps(m[3])), _mm_add_ps(_mm_mul_ps(v32_lo1[1], _mm_set1_ps(m[4])), _mm_mul_ps(v32_lo1[2], _mm_set1_ps(m[5])))), _mm_set1_ps(128.0f));
    ylo2 = _mm_add_ps(_mm_add_ps(_mm_mul_ps(v32_lo2[0], _mm_set1_ps(m[3])), _mm_add_ps(_mm_mul_ps(v32_lo2[1], _mm_set1_ps(m[4])), _mm_mul_ps(v32_lo2[2], _mm_set1_ps(m[5])))), _mm_set1_ps(128.0f));
    yhi1 = _mm_add_ps(_mm_add_ps(_mm_mul_ps(v32_hi1[0], _mm_set1_ps(m[3])), _mm_add_ps(_mm_mul_ps(v32_hi1[1], _mm_set1_ps(m[4])), _mm_mul_ps(v32_hi1[2], _mm_set1_ps(m[5])))), _mm_set1_ps(128.0f));
    yhi2 = _mm_add_ps(_mm_add_ps(_mm_mul_ps(v32_hi2[0], _mm_set1_ps(m[3])), _mm_add_ps(_mm_mul_ps(v32_hi2[1], _mm_set1_ps(m[4])), _mm_mul_ps(v32_hi2[2], _mm_set1_ps(m[5])))), _mm_set1_ps(128.0f));

    ylo1i = _mm_cvtps_epi32(ylo1);
    ylo2i = _mm_cvtps_epi32(ylo2);
    yhi1i = _mm_cvtps_epi32(yhi1);
    yhi2i = _mm_cvtps_epi32(yhi2);

    ylo = _mm_packus_epi32(ylo1i, ylo2i);
    yhi = _mm_packus_epi32(yhi1i, yhi2i);

    yuv[1] = _mm_packus_epi16(ylo, yhi);

    ylo1 = _mm_add_ps(_mm_add_ps(_mm_mul_ps(v32_lo1[0], _mm_set1_ps(m[6])), _mm_add_ps(_mm_mul_ps(v32_lo1[1], _mm_set1_ps(m[7])), _mm_mul_ps(v32_lo1[2], _mm_set1_ps(m[8])))), _mm_set1_ps(128.0f));
    ylo2 = _mm_add_ps(_mm_add_ps(_mm_mul_ps(v32_lo2[0], _mm_set1_ps(m[6])), _mm_add_ps(_mm_mul_ps(v32_lo2[1], _mm_set1_ps(m[7])), _mm_mul_ps(v32_lo2[2], _mm_set1_ps(m[8])))), _mm_set1_ps(128.0f));
    yhi1 = _mm_add_ps(_mm_add_ps(_mm_mul_ps(v32_hi1[0], _mm_set1_ps(m[6])), _mm_add_ps(_mm_mul_ps(v32_hi1[1], _mm_set1_ps(m[7])), _mm_mul_ps(v32_hi1[2], _mm_set1_ps(m[8])))), _mm_set1_ps(128.0f));
    yhi2 = _mm_add_ps(_mm_add_ps(_mm_mul_ps(v32_hi2[0], _mm_set1_ps(m[6])), _mm_add_ps(_mm_mul_ps(v32_hi2[1], _mm_set1_ps(m[7])), _mm_mul_ps(v32_hi2[2], _mm_set1_ps(m[8])))), _mm_set1_ps(128.0f));

    ylo1i = _mm_cvtps_epi32(ylo1);
    ylo2i = _mm_cvtps_epi32(ylo2);
    yhi1i = _mm_cvtps_epi32(yhi1);
    yhi2i = _mm_cvtps_epi32(yhi2);

    ylo = _mm_packus_epi32(ylo1i, ylo2i);
    yhi = _mm_packus_epi32(yhi1i, yhi2i);

    yuv[2] = _mm_packus_epi16(ylo, yhi);

    _mm_storeu_si128((__m128i*)y,yuv[0]);
    _mm_storeu_si128((__m128i*)u,yuv[1]);
    _mm_storeu_si128((__m128i*)v,yuv[2]);
}
c
intel
sse
simd
intrinsics
EN

回答 2

Stack Overflow用户

回答已采纳

发布于 2014-10-03 12:22:22

这里有一个同时找到Y、U和V的解，并且只使用垂直运算符

为了做到这一点，我首先转移了四个像这样的像素

代码语言：javascript
复制
rgbargbargbargba -> rrrrggggbbbbaaaa

使用带掩码的内部_mm_shuffle_epi8。我对16个像素做了这个操作，然后再把它们转过来。

从…

代码语言：javascript
复制
row[0] : rrrrggggbbbbaaaa
row[1] : rrrrggggbbbbaaaa
row[2] : rrrrggggbbbbaaaa
ro2[3] : rrrrggggbbbbaaaa

至

代码语言：javascript
复制
row[0] : rrrrrrrrrrrrrrrr    
row[1] : gggggggggggggggg    
row[2] : bbbbbbbbbbbbbbbb

这与转换4x4整数矩阵的方式相同，如下所示：

代码语言：javascript
复制
__m128i t0 = _mm_unpacklo_epi32(row[0], row[1]);
__m128i t1 = _mm_unpacklo_epi32(row[2], row[3]);
__m128i t2 = _mm_unpackhi_epi32(row[0], row[1]);
__m128i t3 = _mm_unpackhi_epi32(row[2], row[3]);
row[0] = _mm_unpacklo_epi64(t0, t1);
row[1] = _mm_unpackhi_epi64(t0, t1);
row[2] = _mm_unpacklo_epi64(t2, t3);

现在，我将每一行分为高低两部分，并扩展到如下16位。

代码语言：javascript
复制
__m128i v_lo[3], v_hi[3];
for(int i=0; i<3; i++) {
    v_lo[i] = _mm_unpacklo_epi8(row[i],_mm_setzero_si128());
    v_hi[i] = _mm_unpackhi_epi8(row[i],_mm_setzero_si128());
}

最后，我计算Y，U和V如下：

代码语言：javascript
复制
 short m[9] = {66, 129, 25, -38, -74, 112, 112, -94, -18};
__m128i yuv[3];
for(int i=0; i<3; i++) {
    __m128i yuv_lo, yuv_hi;
    yuv_lo = _mm_add_epi16(_mm_add_epi16(
                   _mm_mullo_epi16(v_lo[0], _mm_set1_epi16(m[3*i+0])),
                   _mm_mullo_epi16(v_lo[1], _mm_set1_epi16(m[3*i+1]))),
                   _mm_mullo_epi16(v_lo[2], _mm_set1_epi16(m[3*i+2])));
    yuv_lo = _mm_add_epi16(yuv_lo, _mm_set1_epi16(128));
    yuv_lo = _mm_srli_epi16(yuv_lo, 8);
    yuv_lo = _mm_add_epi16(yuv_lo, _mm_set1_epi16(16));

    yuv_hi = _mm_add_epi16(_mm_add_epi16(
                   _mm_mullo_epi16(v_hi[0], _mm_set1_epi16(m[3*i+0])),
                   _mm_mullo_epi16(v_hi[1], _mm_set1_epi16(m[3*i+1]))),
                   _mm_mullo_epi16(v_hi[2], _mm_set1_epi16(m[3*i+2])));
    yuv_hi = _mm_add_epi16(yuv_hi, _mm_set1_epi16(128));
    yuv_hi = _mm_srli_epi16(yuv_hi, 8);
    yuv_hi = _mm_add_epi16(yuv_hi, _mm_set1_epi16(16));

    yuv[i] = _mm_packus_epi16(yuv_lo,yuv_hi);
}

有关此代码的工作示例，请参阅我的第一个答案和函数rgba2yuv_SSE。

票数 3
EN

Stack Overflow用户

发布于 2014-10-01 21:11:45

这是一个基于OP和Paul的注释的解决方案，内在的_mm_maddubs_epi16要求第二个参数签名，这是g的129因子的一个问题。然而，我们可以通过这样做来解决这个问题。

代码语言：javascript
复制
y = ((66-64)*r + (129-64)*g + (25-64)*b + -64*a) + (64*r + 64*g + 64*b + 64*a)
  = (2*r + 65*g + -39*b -64*a) + 64(r + g + a)

使用这种方法，我们只需要16位整数，我们就可以在这样的时间计算16个y字节：

请注意，我最初使用的是128，但这会导致自255*((25-128)-128)<-32768以来的溢出。

代码语言：javascript
复制
__m128i yk = _mm_set1_epi32(0xc0d94102); -64,-39,64,2
__m128i y4[4];
for(int i=0; i<4; i++) {
    __m128i a4 = _mm_loadu_si128((__m128i*)&a[16*i]);
    __m128i t1 = _mm_maddubs_epi16(a4, yk);
    __m128i t2 = _mm_maddubs_epi16(a4, _mm_set1_epi8(1));
    t2 = _mm_slli_epi16(t2, 6);  //multiply by 64
    y4[i] = _mm_add_epi16(t1,t2);
}
short tmp[8];
_mm_storeu_si128((__m128i*)tmp, y4[0]);
__m128i y8_lo = _mm_hadd_epi16(y4[0], y4[1]);
__m128i y8_hi = _mm_hadd_epi16(y4[2], y4[3]);

y8_lo = _mm_add_epi16(y8_lo, _mm_set1_epi16(128));
y8_lo = _mm_srli_epi16(y8_lo, 8);
y8_lo = _mm_add_epi16(y8_lo, _mm_set1_epi16(16));

y8_hi = _mm_add_epi16(y8_hi, _mm_set1_epi16(128));
y8_hi = _mm_srli_epi16(y8_hi, 8);
y8_hi = _mm_add_epi16(y8_hi, _mm_set1_epi16(16));

__m128i y16 = _mm_packus_epi16(y8_lo,y8_hi);

下面是显示此功能的代码。我将结果与how to perform rgb yuv conversion in C/C++的公式(修改后)进行了比较，即：

代码语言：javascript
复制
#define CLIP(X) ( (X) > 255 ? 255 : (X) < 0 ? 0 : X)
#define RGB2Y(R, G, B) CLIP(( (  66 * (0xff & R) + 129 * (0xff & G) +  25 * (0xff & B) + 128) >> 8) +  16)

守则：

代码语言：javascript
复制
#include <stdio.h>
#include <x86intrin.h>
#include <stdlib.h>

#define CLIP(X) ( (X) > 255 ? 255 : (X) < 0 ? 0 : X)
#define RGB2Y(R, G, B) CLIP(( (  66 * (0xff & R) + 129 * (0xff & G) +  25 * (0xff & B) + 128) >> 8) +  16)

void rgba2y_SSE_v1(char *a, char *b) {
    __m128i yk = _mm_setr_epi16(66,129,25,0, 66,129,25,0);
    __m128i out[4];
    for(int i=0; i<4; i++) {        
        __m128i a4, lo, hi;
        a4 = _mm_loadu_si128((__m128i*)&a[16*i]);
        lo = _mm_unpacklo_epi8(a4,_mm_setzero_si128());
        hi = _mm_unpackhi_epi8(a4,_mm_setzero_si128());

        lo = _mm_madd_epi16(lo,yk);
        lo = _mm_hadd_epi32(lo,lo);

        hi  = _mm_madd_epi16(hi,yk);
        hi  = _mm_hadd_epi32(hi,hi);
        out[i] = _mm_unpackhi_epi64(lo,hi);
    }
    __m128i out_lo = _mm_packus_epi32(out[0], out[1]);
    __m128i out_hi = _mm_packus_epi32(out[2], out[3]);

    out_lo = _mm_add_epi16(out_lo, _mm_set1_epi16(128));
    out_lo = _mm_srli_epi16(out_lo, 8);
    out_lo = _mm_add_epi16(out_lo, _mm_set1_epi16(16)); 

    out_hi = _mm_add_epi16(out_hi, _mm_set1_epi16(128));
    out_hi = _mm_srli_epi16(out_hi, 8);
    out_hi = _mm_add_epi16(out_hi, _mm_set1_epi16(16)); 

    __m128i y16 = _mm_packus_epi16(out_lo,out_hi);
    _mm_storeu_si128((__m128i*)b,y16);
}

void rgba2y_SSE_v2(char *a, char *b) {
    __m128i yk = _mm_set1_epi32(0xc0d94102);
    __m128i y4[4];
    for(int i=0; i<4; i++) {
        __m128i a4 = _mm_loadu_si128((__m128i*)&a[16*i]);
        __m128i t1 = _mm_maddubs_epi16(a4, yk);
        __m128i t2 = _mm_maddubs_epi16(a4, _mm_set1_epi8(1));
        t2 = _mm_slli_epi16(t2, 6);
        y4[i] = _mm_add_epi16(t1,t2); 
    } 
    short tmp[8];
    _mm_storeu_si128((__m128i*)tmp, y4[0]);
    __m128i y8_lo = _mm_hadd_epi16(y4[0], y4[1]);
    __m128i y8_hi = _mm_hadd_epi16(y4[2], y4[3]);

    y8_lo = _mm_add_epi16(y8_lo, _mm_set1_epi16(128));
    y8_lo = _mm_srli_epi16(y8_lo, 8);
    y8_lo = _mm_add_epi16(y8_lo, _mm_set1_epi16(16)); 

    y8_hi = _mm_add_epi16(y8_hi, _mm_set1_epi16(128));
    y8_hi = _mm_srli_epi16(y8_hi, 8);
    y8_hi = _mm_add_epi16(y8_hi, _mm_set1_epi16(16)); 

    __m128i y16 = _mm_packus_epi16(y8_lo,y8_hi);
    _mm_storeu_si128((__m128i*)b,y16);
}

void rgba2yuv_SSE(char *a, char *b) {
    __m128i mask = _mm_setr_epi8(0x00,0x04,0x08,0x0c, 0x01,0x05,0x09,0x0d, 0x02,0x06,0x0a,0x0e, 0x03,0x07,0x0b,0x0f);
    short m[9] = {66, 129, 25, -38, -74, 112, 112, -94, -18};

    __m128i row[4];
    for(int i=0; i<4; i++) {
        row[i] = _mm_loadu_si128((__m128i*)&a[16*i]);
        row[i] = _mm_shuffle_epi8(row[i],mask);
    }

    __m128i t0 = _mm_unpacklo_epi32(row[0], row[1]);
    __m128i t1 = _mm_unpacklo_epi32(row[2], row[3]);
    __m128i t2 = _mm_unpackhi_epi32(row[0], row[1]);
    __m128i t3 = _mm_unpackhi_epi32(row[2], row[3]);
    row[0] = _mm_unpacklo_epi64(t0, t1);
    row[1] = _mm_unpackhi_epi64(t0, t1);
    row[2] = _mm_unpacklo_epi64(t2, t3);

    __m128i v_lo[3], v_hi[3];
    for(int i=0; i<3; i++) {
        v_lo[i] = _mm_unpacklo_epi8(row[i],_mm_setzero_si128());
        v_hi[i] = _mm_unpackhi_epi8(row[i],_mm_setzero_si128());
    }

    __m128i yuv[3];
    for(int i=0; i<3; i++) {
        __m128i yuv_lo, yuv_hi;
        yuv_lo = _mm_add_epi16(_mm_add_epi16(
                       _mm_mullo_epi16(v_lo[0], _mm_set1_epi16(m[3*i+0])),
                       _mm_mullo_epi16(v_lo[1], _mm_set1_epi16(m[3*i+1]))),
                       _mm_mullo_epi16(v_lo[2], _mm_set1_epi16(m[3*i+2])));
        yuv_lo = _mm_add_epi16(yuv_lo, _mm_set1_epi16(128));
        yuv_lo = _mm_srli_epi16(yuv_lo, 8);
        yuv_lo = _mm_add_epi16(yuv_lo, _mm_set1_epi16(16)); 

        yuv_hi = _mm_add_epi16(_mm_add_epi16(
                       _mm_mullo_epi16(v_hi[0], _mm_set1_epi16(m[3*i+0])),
                       _mm_mullo_epi16(v_hi[1], _mm_set1_epi16(m[3*i+1]))),
                       _mm_mullo_epi16(v_hi[2], _mm_set1_epi16(m[3*i+2])));
        yuv_hi = _mm_add_epi16(yuv_hi, _mm_set1_epi16(128));
        yuv_hi = _mm_srli_epi16(yuv_hi, 8);
        yuv_hi = _mm_add_epi16(yuv_hi, _mm_set1_epi16(16)); 

        yuv[i] = _mm_packus_epi16(yuv_lo,yuv_hi);
    }   
    _mm_storeu_si128((__m128i*)b,yuv[0]);
}



int main(void) {
    char rgba[64];
    char y1[16], y2[16], yuv[48];
    for(int i=0; i<64; i++) rgba[i] = rand()%256;
    rgba2y_SSE_v1(rgba,y1);
    rgba2y_SSE_v2(rgba,y2);
    rgba2yuv_SSE(rgba,yuv);

    printf("RGB2Y: "); for(int i=0; i<16; i++) printf("%x ", 0xff & RGB2Y(rgba[4*i+0], rgba[4*i+1], rgba[4*i+2])); printf("\n");
    printf("SSE_v1 "); for(int i=0; i<16; i++) printf("%x ", 0xff & y1[i]); printf("\n");
    printf("SSE_v2 "); for(int i=0; i<16; i++) printf("%x ", 0xff & y2[i]); printf("\n");
    printf("SSE_v3 "); for(int i=0; i<16; i++) printf("%x ", 0xff & yuv[i]); printf("\n");

}

产出：

代码语言：javascript
复制
RGB2Y: 99 ad 94 e3 9a a2 60 81 45 59 49 a5 aa 9b 60 4d 
SSE_v1 99 ad 94 e3 9a a2 60 81 45 59 49 a5 aa 9b 60 4d 
SSE_v2 99 ad 94 e3 9a a2 60 81 45 59 49 a5 aa 9b 60 4d 
SSE_v3 99 ad 94 e3 9a a2 60 81 45 59 49 a5 aa 9b 60 4d
票数 3
EN
页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：

https://stackoverflow.com/questions/26096265

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