为什么我简单的四元数乘法比SSE快? [英] Why is my straightforward quaternion multiplication faster than SSE?
问题描述
我一直在研究几种不同的四元数乘法实现，但是我很惊讶地看到参考实现到目前为止是我最快的.这是有问题的实现:
I've been going through a few different quaternion multiplication implementations, but I've been rather surprised to see that the reference implementation is, so far, my fastest. This is the implementation in question:
inline static quat multiply(const quat& lhs, const quat& rhs)
{
return quat((lhs.w * rhs.x) + (lhs.x * rhs.w) + (lhs.y * rhs.z) - (lhs.z * rhs.y),
(lhs.w * rhs.y) + (lhs.y * rhs.w) + (lhs.z * rhs.x) - (lhs.x * rhs.z),
(lhs.w * rhs.z) + (lhs.z * rhs.w) + (lhs.x * rhs.y) - (lhs.y * rhs.x),
(lhs.w * rhs.w) - (lhs.x * rhs.x) - (lhs.y * rhs.y) - (lhs.z * rhs.z));
}
我已经尝试了其他一些实现，其中一些使用SSE，而有些则没有.这是一个这样的SSE实现示例，基本上是从Bullet Physics使用的库中复制的:
I've tried a few other implementations, some using SSE, some that don't. Here's an example of one such SSE implementation, basically copied from the library that Bullet Physics uses:
inline static __m128 multiplynew(__m128 lhs, __m128 rhs)
{
__m128 qv, tmp0, tmp1, tmp2, tmp3;
__m128 product, l_wxyz, r_wxyz, xy, qw;
vec4 sw;
tmp0 = _mm_shuffle_ps(lhs, lhs, _MM_SHUFFLE(3, 0, 2, 1));
tmp1 = _mm_shuffle_ps(rhs, rhs, _MM_SHUFFLE(3, 1, 0, 2));
tmp2 = _mm_shuffle_ps(lhs, lhs, _MM_SHUFFLE(3, 1, 0, 2));
tmp3 = _mm_shuffle_ps(rhs, rhs, _MM_SHUFFLE(3, 0, 2, 1));
qv = _mm_mul_ps(_mm_splat_ps(lhs, 3), rhs);
qv = _mm_madd_ps(_mm_splat_ps(rhs, 3), lhs, qv);
qv = _mm_madd_ps(tmp0, tmp1, qv);
qv = _mm_nmsub_ps(tmp2, tmp3, qv);
product = _mm_mul_ps(lhs, rhs);
l_wxyz = _mm_sld_ps(lhs, lhs, 12);
r_wxyz = _mm_sld_ps(rhs, rhs, 12);
qw = _mm_nmsub_ps(l_wxyz, r_wxyz, product);
xy = _mm_madd_ps(l_wxyz, r_wxyz, product);
qw = _mm_sub_ps(qw, _mm_sld_ps(xy, xy, 8));
sw.uiw = 0xffffffff;
return _mm_sel_ps(qv, qw, sw);
}
在启用了优化的发布模式下，我的简单参考实现比bullet的SSE实现快70％-90％.在没有优化的调试模式下，它的运行速度快了3倍.
In release mode with optimizations turned on, my simple reference implementation runs 70%-90% faster than bullet's SSE implementation. In debug mode without optimizations, it runs as much as 3x faster.
我的第一个问题是，为什么会这样?
My first question is, why does this happen?
我的第二个问题是，我有什么方法可以优化四元数-四元数乘法例程?我不想处理汇编，但是我在其他地方经常使用sse内在函数.
My second question is, is there any way for me to optimize my quaternion-quaternion multiplication routine? I don't want to deal with assembly, but I use sse intrinsics quite a lot elsewhere.
(顺便说一句，我的四元数的数据存储定义为union { __m128 data; struct { float x, y, z, w; }; float f[4]; };)
(btw, if it matters, the data storage of my quaternion is defined as union { __m128 data; struct { float x, y, z, w; }; float f[4]; };)
我看了看反汇编.这是multiply(快速的非sse代码)的反汇编:
I looked at the dissassembly. Here's the disassembly for multiply (the fast non-sse one):
00EC9940 movaps xmm3,xmmword ptr [esp+0D0h]
00EC9948 movaps xmm2,xmmword ptr [esp+0C0h]
00EC9950 movaps xmm4,xmm3
00EC9953 mulss xmm4,xmm5
00EC9957 movaps xmm0,xmm2
00EC995A mulss xmm0,xmm6
00EC995E mulss xmm3,xmm1
00EC9962 addss xmm4,xmm0
00EC9966 movss xmm0,dword ptr [esp+40h]
00EC996C mulss xmm0,xmm1
00EC9970 addss xmm4,xmm0
00EC9974 movss xmm0,dword ptr [esp+0F0h]
00EC997D mulss xmm0,xmm7
00EC9981 subss xmm4,xmm0
00EC9985 movss xmm0,dword ptr [esp+0F0h]
00EC998E mulss xmm0,xmm6
00EC9992 addss xmm3,xmm0
00EC9996 movaps xmm0,xmm2
00EC9999 movaps xmm2,xmmword ptr [esp+40h]
00EC999E mulss xmm0,xmm7
00EC99A2 addss xmm3,xmm0
00EC99A6 movaps xmm0,xmm2
00EC99A9 mulss xmm0,xmm5
00EC99AD mulss xmm2,xmm6
00EC99B1 subss xmm3,xmm0
00EC99B5 movss xmm0,dword ptr [esp+0D0h]
00EC99BE mulss xmm0,xmm7
00EC99C2 addss xmm2,xmm0
00EC99C6 movss xmm0,dword ptr [esp+0F0h]
00EC99CF mulss xmm0,xmm5
00EC99D3 addss xmm2,xmm0
00EC99D7 movss xmm0,dword ptr [esp+0C0h]
00EC99E0 mulss xmm0,xmm1
00EC99E4 movss xmm1,dword ptr [esp+0D0h]
00EC99ED mulss xmm1,xmm6
00EC99F1 subss xmm2,xmm0
00EC99F5 movss xmm0,dword ptr [esp+0C0h]
00EC99FE mulss xmm0,xmm5
00EC9A02 movaps xmm5,xmmword ptr [esp+50h]
00EC9A07 unpcklps xmm4,xmm2
00EC9A0A subss xmm1,xmm0
00EC9A0E movss xmm0,dword ptr [esp+0F0h]
00EC9A17 mulss xmm0,xmm5
00EC9A1B subss xmm1,xmm0
00EC9A1F movss xmm0,dword ptr [esp+40h]
00EC9A25 mulss xmm0,xmm7
00EC9A29 subss xmm1,xmm0
00EC9A2D unpcklps xmm3,xmm1
00EC9A30 unpcklps xmm4,xmm3
00EC9A33 movaps xmm5,xmm4
00EC9A36 movaps xmmword ptr [esp+30h],xmm5
00EC9A3B dec eax
00EC9A3C je SDL_main+58Ah (0EC9A5Ah)
这是multiplynew(慢的sse之一)的反汇编:
And here's the disassembly for multiplynew (the slow sse one):
00329BF3 movaps xmm6,xmm5
00329BF6 mulps xmm6,xmm1
00329BF9 movaps xmm0,xmm5
00329BFC mov dword ptr [esp+6Ch],0FFFFFFFFh
00329C04 shufps xmm0,xmm5,93h
00329C08 movaps xmm1,xmm5
00329C0B mulps xmm4,xmm0
00329C0E movaps xmm0,xmmword ptr [esp+110h]
00329C16 movaps xmm3,xmm6
00329C19 shufps xmm1,xmm5,0FFh
00329C1D mulps xmm1,xmmword ptr [esp+40h]
00329C22 movaps xmm7,xmmword ptr [esp+60h]
00329C27 addps xmm3,xmm4
00329C2A mulps xmm0,xmm5
00329C2D subps xmm6,xmm4
00329C30 shufps xmm3,xmm3,4Eh
00329C34 addps xmm1,xmm0
00329C37 movaps xmm0,xmm5
00329C3A shufps xmm0,xmm5,0C9h
00329C3E subps xmm6,xmm3
00329C41 mulps xmm0,xmmword ptr [esp+120h]
00329C49 shufps xmm5,xmm5,0D2h
00329C4D mulps xmm5,xmmword ptr [esp+0C0h]
00329C55 andps xmm6,xmmword ptr [esp+60h]
00329C5A addps xmm1,xmm0
00329C5D subps xmm1,xmm5
00329C60 andnps xmm7,xmm1
我测试速度的方法是使用:
The way I test the speed is using:
timer.update();
for (uint i = 0; i < 1000000; ++i)
{
temp1 = quat::multiply(temp1, q1);
}
timer.update();
printf("1M calls to multiplyOld took %fs.\n", timer.getDeltaTime());
(timer.getDeltaTime()返回上次调用timer.update()与在此之前调用timer.update()的时间之间经过的时间(以秒为单位).
(timer.getDeltaTime() returns how much time has passed, in seconds, between the last time timer.update() was called and the time that timer.update() was called before that.)
为什么我的非SSE版本运行速度更快，尽管有更多说明.我是在阅读反汇编中的错误信息吗?
Why does my non-sse version run faster despite having more instructions..? Am I reading the disassembly wrong or something?
我发现在x64中进行编译时，sse版本的运行速度比非sse版本的运行速度快.
I've discovered that the sse version is running faster than the non-sse version when I compile in x64.
推荐答案
使用SIMD的最佳方法是使用SSE(AVX)一次乘以四个(八个)独立的四元数.但是，这通常是不切实际的.在这种情况下，我建议您查看Agner Fog的 vectorclass .在目录special中，他有一个文件quaterinon.h.我转换了乘法函数以匹配您的代码.这仅需要SSE2(#include <emmintrin.h>).
The best way to use SIMD would be to multiply four (eight) independent quaternions at once with SSE (AVX). However, this often is not practical. In that case I recommend checking out Agner Fog's vectorclass. In the directory special he has a file quaterinon.h. I converted the multiply function to match your code. This only needs SSE2 (#include <emmintrin.h>).
inline static __m128 multiplynew(__m128 a, __m128 b) {
__m128 a1123 = _mm_shuffle_ps(a,a,0xE5);
__m128 a2231 = _mm_shuffle_ps(a,a,0x7A);
__m128 b1000 = _mm_shuffle_ps(b,b,0x01);
__m128 b2312 = _mm_shuffle_ps(b,b,0x9E);
__m128 t1 = _mm_mul_ps(a1123, b1000);
__m128 t2 = _mm_mul_ps(a2231, b2312);
__m128 t12 = _mm_add_ps(t1, t2);
const __m128i mask =_mm_set_epi32(0,0,0,0x80000000);
__m128 t12m = _mm_xor_ps(t12, _mm_castsi128_ps(mask)); // flip sign bits
__m128 a3312 = _mm_shuffle_ps(a,a,0x9F);
__m128 b3231 = _mm_shuffle_ps(b,b,0x7B);
__m128 a0000 = _mm_shuffle_ps(a,a,0x00);
__m128 t3 = _mm_mul_ps(a3312, b3231);
__m128 t0 = _mm_mul_ps(a0000, b);
__m128 t03 = _mm_sub_ps(t0, t3);
return _mm_add_ps(t03, t12m);
}
这是程序集输出(GCC 4.8 MASM样式):
Here is the assembly output (GCC 4.8 MASM style):
multiplynew(float __vector, float __vector):
movaps xmm2, xmm0
movaps xmm3, xmm0
movaps xmm5, xmm1
movaps xmm4, xmm1
shufps xmm2, xmm0, 229
shufps xmm4, xmm1, 158
shufps xmm3, xmm0, 122
shufps xmm5, xmm1, 1
mulps xmm3, xmm4
movaps xmm4, xmm1
mulps xmm2, xmm5
shufps xmm4, xmm1, 123
addps xmm2, xmm3
movaps xmm3, xmm0
shufps xmm3, xmm0, 159
shufps xmm0, xmm0, 0
xorps xmm2, XMMWORD PTR .LC0[rip]
mulps xmm3, xmm4
mulps xmm0, xmm1
subps xmm0, xmm3
addps xmm0, xmm2
ret
这篇关于为什么我简单的四元数乘法比SSE快?的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！
