((c1a * c2a) - (c1b * c2b)) + ((c1b * c2a) + (c1a * c2b))i

所以你的2个组件是一个复数

((c1a * c2a) - (c1b * c2b)) and ((c1b * c2a) + (c1a * c2b))i

因此,假设您使用8个浮点数来表示如下定义的4个复数：

c1a,c1b,c2a,c2b
c3a,c3b,c4a,c4b

并且你想同时做(c1 * c3)和(c2 * c4)你的SSE代码看起来像“下面的东西”：

(注意我在windows下使用了MSVC,但原理是相同的).

__declspec( align( 16 ) ) float c1c2[]        = { 1.0f,2.0f,3.0f,4.0f };
__declspec( align( 16 ) ) float c3c4[]          = { 4.0f,1.0f };
__declspec( align( 16 ) ) float mulfactors[]    = { -1.0f,1.0f,-1.0f,1.0f };
__declspec( align( 16 ) ) float res[]           = { 0.0f,0.0f,0.0f };

__asm 
{
    movaps xmm0,xmmword ptr [c1c2]         // Load c1 and c2 into xmm0.
    movaps xmm1,xmmword ptr [c3c4]         // Load c3 and c4 into xmm1.
    movaps xmm4,xmmword ptr [mulfactors]   // load multiplication factors into xmm4

    movaps xmm2,xmm1                       
    movaps xmm3,xmm0                       
    shufps xmm2,xmm1,0xA0                 // Change order to c3a c3a c4a c4a and store in xmm2
    shufps xmm1,0xF5                 // Change order to c3b c3b c4b c4b and store in xmm1
    shufps xmm3,xmm0,0xB1                 // change order to c1b c1a c2b c2a abd store in xmm3

    mulps xmm0,xmm2                        
    mulps xmm3,xmm1                    
    mulps xmm3,xmm4                        // Flip the signs of the 'a's so the add works correctly.

    addps xmm0,xmm3                        // Add together

    movaps xmmword ptr [res],xmm0          // Store back out
};

float res1a = (c1c2[0] * c3c4[0]) - (c1c2[1] * c3c4[1]);
float res1b = (c1c2[1] * c3c4[0]) + (c1c2[0] * c3c4[1]);

float res2a = (c1c2[2] * c3c4[2]) - (c1c2[3] * c3c4[3]);
float res2b = (c1c2[3] * c3c4[2]) + (c1c2[2] * c3c4[3]);

if ( res1a != res[0] || 
     res1b != res[1] || 
     res2a != res[2] || 
     res2b != res[3] )
{
    _exit( 1 );
}

我上面所做的是我将数学简化了一下.假设如下：

c1a c1b c2a c2b
c3a c3b c4a c4b

通过重新排列我最终得到以下向量

0 => c1a c1b c2a c2b
1 => c3b c3b c4b c4b
2 => c3a c3a c4a c4a
3 => c1b c1a c2b c2a

然后我将0和2相乘得到：

0 => c1a * c3a,c1b * c3a,c2a * c4a,c2b * c4a

接下来我将3和1相乘得到：

3 => c1b * c3b,c1a * c3b,c2b * c4b,c2a * c4b

最后,我在3中翻转了几个花车的标志

3 => -(c1b * c3b),-(c2b * c4b),c2a * c4b

所以我可以把它们加在一起然后得到

(c1a * c3a) - (c1b * c3b),(c1b * c3a ) + (c1a * c3b),(c2a * c4a) - (c2b * c4b),(c2b * c4a) + (c2a * c4b)

这是我们之后的:)