复杂的Mul和Div使用sse说明

Question

通过SSE指令执行复杂的乘法和除法是否有益？我知道使用SSE时加法和减法表现更好.有人能告诉我如何使用SSE执行复杂的乘法以获得更好的性能吗？

Answer 1

为了完整起见,可在此处下载的英特尔®64和IA-32架构优化参考手册包含复数乘法(示例6-9)和复数除法(示例6-10)的汇编.

这是例如乘法代码:

// Multiplication of (ak + i bk ) * (ck + i dk )
// a + i b can be stored as a data structure
movsldup xmm0, src1; load real parts into the destination, a1, a1, a0, a0
movaps xmm1, src2; load the 2nd pair of complex values, i.e. d1, c1, d0, c0
mulps xmm0, xmm1; temporary results, a1d1, a1c1, a0d0, a0c0
shufps xmm1, xmm1, b1; reorder the real and imaginary parts, c1, d1, c0, d0
movshdup xmm2, src1; load imaginary parts into the destination, b1, b1, b0, b0
mulps xmm2, xmm1; temporary results, b1c1, b1d1, b0c0, b0d0
addsubps xmm0, xmm2; b1c1+a1d1, a1c1 -b1d1, b0c0+a0d0, ; a0c0-b0d0

程序集直接映射到gccs X86内在函数(只是用每个指令作谓词__builtin_ia32_).

Answer 2

复杂的乘法定义为:

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

所以你的2个组件就是一个复数

((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, 3.0f, 2.0f, 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, 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, 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), c1a * 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)

这是我们追求的:)