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，可下载here的英特尔®64和IA-32架构优化参考手册包含复数乘法（例6-9）和复数除法（例6-10）的汇编。

这是例如乘法码：

// 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 intrinsics（只使用__builtin_ia32_预测每条指令。）

Answer 2

复杂的乘法定义为：

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

因此，复数中的2个组件将是

((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

你想同时做（c1 * c3）和（c2 * c4）你的SSE代码看起来像“以下”：

（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，我翻转了3个浮标的迹象

3 => -(c1b * c3b), c1a * c3b, -(c2b * c4b), c2a * c4b

所以我可以将它们添加到一起并获得

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

我们追求的是：）

Answer 3

英特尔优化参考中的算法无法正确处理输入中的溢出和NaN。

数字的实部或虚部中的单个NaN将错误地传播到另一部分。

由于无穷大的几个操作（例如，无穷大* 0）以NaN结尾，溢出可能会导致NaN出现在您的行为良好的数据中。

如果溢出和NaN很少见，一种避免这种情况的简单方法是只检查结果中的NaN并使用编译器IEEE兼容的实现重新计算它：

float complex a[2], b[2];
__m128 res = simd_fast_multiply(a, b);

/* store unconditionally, can be executed in parallel with the check
 * making it almost free if there is no NaN in data */
_mm_store_ps(dest, res);

/* check for NaN */
__m128 n = _mm_cmpneq_ps(res, res);
int have_nan = _mm_movemask_ps(n);
if (have_nan != 0) {
    /* do it again unvectorized */
    dest[0] = a[0] * b[0];
    dest[1] = a[1] * b[1];
}

复杂的Mul和Div使用sse说明

3 个答案: