我想用Eigen编写一些上证所代码,有些行为没有我。上证所表现特征
鉴于代码:
#ifndef EIGEN_DONT_VECTORIZE // Not needed with Intel C++ Compiler XE 15.0
#define EIGEN_VECTORIZE_SSE4_2
#define EIGEN_VECTORIZE_SSE4_1
#define EIGEN_VECTORIZE_SSSE3
#define EIGEN_VECTORIZE_SSE3
#endif
#include "stdafx.h"
#include <iostream>
#include <unsupported/Eigen/AlignedVector3>
#include <Eigen/StdVector>
#include <chrono>
int _tmain(int argc, _TCHAR* argv[]) {
static const int SIZE = 4000000;
EIGEN_ALIGNED_VECTOR3 Eigen::AlignedVector3<float> A_SSE(1, 1, 1);
//EIGEN_ALIGNED_VECTOR3 Eigen::AlignedVector3<float> B_SSE(2, 2, 2);
//std::vector<Eigen::AlignedVector3<float>> C_SSE(SIZE, Eigen::AlignedVector3<float>(0,0,0));
EIGEN_ALIGNED_VECTOR3 Eigen::AlignedVector3<float> A_SSE1(1, 1, 1);
EIGEN_ALIGNED_VECTOR3 Eigen::AlignedVector3<float> A_SSE2(1, 1, 1);
EIGEN_ALIGNED_VECTOR3 Eigen::AlignedVector3<float> A_SSE3(1, 1, 1);
EIGEN_ALIGNED_VECTOR3 Eigen::AlignedVector3<float> A_SSE4(1, 1, 1);
EIGEN_ALIGNED_VECTOR3 Eigen::AlignedVector3<float> B_SSE(2, 2, 2);
EIGEN_ALIGNED_VECTOR3 Eigen::AlignedVector3<float> B_SSE_increment_unroll(16, 16, 16);
A_SSE2 += B_SSE;
A_SSE3 = A_SSE2 + B_SSE;
A_SSE4 = A_SSE3 + B_SSE;
std::vector<Eigen::AlignedVector3<float>> C_SSE(SIZE, Eigen::AlignedVector3<float>(0, 0, 0));
auto start2 = std::chrono::system_clock::now();
// no unroll
for (int iteration = 0; iteration < SIZE; ++iteration) {
A_SSE += B_SSE;
C_SSE[iteration] = A_SSE;
}
//// own unroll
//for (int iteration = 0; iteration < SIZE/8; ++iteration){
// A_SSE1 += B_SSE_increment_unroll;
// A_SSE2 += B_SSE_increment_unroll;
// A_SSE3 += B_SSE_increment_unroll;
// A_SSE4 += B_SSE_increment_unroll;
// C_SSE[iteration * 2] = A_SSE1;
// C_SSE[iteration * 2 + 1] = A_SSE2;
// C_SSE[iteration * 2 + 2] = A_SSE3;
// C_SSE[iteration * 2 + 3] = A_SSE4;
//}
auto end2 = std::chrono::system_clock::now();
auto elapsed2 = end2 - start2;
std::cout << "Eigen aligned vector " << elapsed2.count() << '\n';
Eigen::Matrix3Xf A = Eigen::Matrix3Xf::Zero(3, SIZE);
Eigen::Vector3f B(3, 3, 3);
Eigen::Vector3f C(2, 2, 2);
auto start1 = std::chrono::system_clock::now();
for (int iteration = 0; iteration < SIZE; ++iteration) {
B += C;
A.col(iteration) = B;
}
auto end1 = std::chrono::system_clock::now();
auto elapsed1 = end1 - start1;
std::cout << "Eigen matrix " << elapsed1.count() << '\n';
float *pResult = (float*)_aligned_malloc(SIZE * sizeof(float) * 4, 16); // align to 16-byte for SSE
auto start3 = std::chrono::system_clock::now();
__m128 x;
__m128 xDelta = _mm_set1_ps(2.0f); // Set the xDelta to (4,4,4,4)
__m128 *pResultSSE = (__m128*) pResult;
x = _mm_set_ps(1.0f, 1.0f, 1.0f, 1.0f); // Set the initial values of x to (4,3,2,1)
for (int iteration = 0; iteration < SIZE; ++iteration)
{
x = _mm_add_ps(x, xDelta);
pResultSSE[iteration] = x;
}
auto end3 = std::chrono::system_clock::now();
auto elapsed3 = end3 - start3;
std::cout << "Own sse " << elapsed3.count() << '\n';
}
时机似乎很奇怪,在我的电脑
- 征对准矢量展开:20057
- 征对齐矢量没有UNROLL:〜120320
- 特征矩阵: 〜120207(与Align不展开相同)
- 自己的SSE:160784
当我检查程序集,对齐版本和Own SSE时,使用addps movaps,但是直到我手动展开循环,我没有获得额外的性能,即使我没有在所有运行中执行(50%),没有任何提升。版本机智Eigen Matrix不使用sse,实现相同的性能,内联汇编显示在16次迭代中展开。手动展开是否有影响力?我们是否应该为SSE手动执行此操作,并且如果使用它的CPU属性取决于它?
编辑: 所以总结一下。由于无法证明展开循环与未展开的展开循环相同,SSE指令执行效果不佳,因此无法隐藏存储器存储延迟。但是在汇编代码中,“单个”指令只使用1个寄存器并在展开的循环中递增。如果SSE上瘾是垂直执行的(对齐向量中的单个浮点积累了相同的添加操作量),编译器应该能够证明展开的平等性。默认情况下,SSE操作是否未经编译器优化?如果展开循环保持执行顺序,那么保留非关联数学运算,自动展开应该是可能的,为什么它不会发生,以及如何强制编译器执行它?
编辑: 作为建议我跑的测试,但是从本征替补单位不下的Visual Studio 2017年因此被
#include <iostream>
#include <vector>
#include <unsupported/Eigen/AlignedVector3>
#include <chrono>
#include <numeric>
EIGEN_DONT_INLINE
void vector_no_unroll(std::vector<Eigen::AlignedVector3<float>>& out)
{
Eigen::AlignedVector3<float> A_SSE(1, 1, 1);
Eigen::AlignedVector3<float> B_SSE(2, 2, 2);
for (auto &x : out)
{
A_SSE += B_SSE;
x = A_SSE;
}
}
EIGEN_DONT_INLINE
void vector_unrolled(std::vector<Eigen::AlignedVector3<float>>& out)
{
Eigen::AlignedVector3<float> A_SSE1(1, 1, 1);
Eigen::AlignedVector3<float> A_SSE2(1, 1, 1);
Eigen::AlignedVector3<float> A_SSE3(1, 1, 1);
Eigen::AlignedVector3<float> A_SSE4(1, 1, 1);
Eigen::AlignedVector3<float> B_SSE(2, 2, 2);
Eigen::AlignedVector3<float> B_SSE_increment_unroll(16, 16, 16);
A_SSE2 += B_SSE;
A_SSE3 = A_SSE2 + B_SSE;
A_SSE4 = A_SSE3 + B_SSE;
for (size_t i = 0; i<out.size(); i += 4)
{
A_SSE1 += B_SSE_increment_unroll;
A_SSE2 += B_SSE_increment_unroll;
A_SSE3 += B_SSE_increment_unroll;
A_SSE4 += B_SSE_increment_unroll;
out[i + 0] = A_SSE1;
out[i + 1] = A_SSE2;
out[i + 2] = A_SSE3;
out[i + 3] = A_SSE4;
}
}
EIGEN_DONT_INLINE
void eigen_matrix(Eigen::Matrix3Xf& out)
{
Eigen::Vector3f B(1, 1, 1);
Eigen::Vector3f C(2, 2, 2);
for (int i = 0; i < out.cols(); ++i) {
B += C;
out.col(i) = B;
}
}
template<int unrolling> EIGEN_DONT_INLINE
void eigen_matrix_unrolled(Eigen::Matrix3Xf& out)
{
Eigen::Matrix<float, 3, unrolling> B = Eigen::Matrix<float, 1, unrolling>::LinSpaced(3.f, 1 + 2 * unrolling).template replicate<3, 1>();
for (int i = 0; i < out.cols(); i += unrolling) {
out.middleCols<unrolling>(i) = B;
B.array() += float(2 * unrolling);
}
}
int main() {
static const int SIZE = 4000000;
int tries = 30;
int rep = 10;
std::vector<int> Timings(tries, 0);
{
Eigen::Matrix3Xf A(3, SIZE);
#pragma loop(1)
for (int iter = 0; iter < tries; ++iter)
{
auto start1 = std::chrono::system_clock::now();
eigen_matrix(A);
Timings[iter] = (std::chrono::system_clock::now() - start1).count();
}
}
std::cout << "eigen matrix Min: " << *std::min_element(Timings.begin(), Timings.end()) << " ms\n";
std::cout << "eigen matrix Mean: " << std::accumulate(Timings.begin(), Timings.end(), 0)/tries << " ms\n";
{
Eigen::Matrix3Xf A(3, SIZE);
#pragma loop(1)
for (int iter = 0; iter < tries; ++iter)
{
auto start1 = std::chrono::system_clock::now();
eigen_matrix_unrolled<4>(A);
Timings[iter] = (std::chrono::system_clock::now() - start1).count();
}
}
std::cout << "eigen matrix unrolled 4 min: " << *std::min_element(Timings.begin(), Timings.end()) << " ms\n";
std::cout << "eigen matrix unrolled 4 Mean: " << std::accumulate(Timings.begin(), Timings.end(), 0)/tries << " ms\n";
{
Eigen::Matrix3Xf A(3, SIZE);
#pragma loop(1)
for (int iter = 0; iter < tries; ++iter)
{
auto start1 = std::chrono::system_clock::now();
eigen_matrix_unrolled<8>(A);
Timings[iter] = (std::chrono::system_clock::now() - start1).count();
}
}
std::cout << "eigen matrix unrolled 8 min: " << *std::min_element(Timings.begin(), Timings.end()) << " ms\n";
std::cout << "eigen matrix unrolled 8 Mean: " << std::accumulate(Timings.begin(), Timings.end(), 0)/tries << " ms\n";
{
std::vector<Eigen::AlignedVector3<float>> A(SIZE, Eigen::AlignedVector3<float>(0, 0, 0));
#pragma loop(1)
for (int iter = 0; iter < tries; ++iter)
{
auto start1 = std::chrono::system_clock::now();
vector_no_unroll(A);
Timings[iter] = (std::chrono::system_clock::now() - start1).count();
}
}
std::cout << "eigen vector min: " << *std::min_element(Timings.begin(), Timings.end()) << " ms\n";
std::cout << "eigen vector Mean: " << std::accumulate(Timings.begin(), Timings.end(), 0)/tries << " ms\n";
{
std::vector<Eigen::AlignedVector3<float>> A(SIZE, Eigen::AlignedVector3<float>(0, 0, 0));
#pragma loop(1)
for (int iter = 0; iter < tries; ++iter)
{
auto start1 = std::chrono::system_clock::now();
vector_unrolled(A);
Timings[iter] = (std::chrono::system_clock::now() - start1).count();
}
}
std::cout << "eigen vector unrolled min: " << *std::min_element(Timings.begin(), Timings.end()) << " ms\n";
std::cout << "eigen vector unrolled Mean: " << std::accumulate(Timings.begin(), Timings.end(), 0)/tries << " ms\n";
}
更换工作,并检查结果在8个指出错误机(所有窗口)和得到如下结果
特征矩阵民:110477毫秒
特征矩阵平均值:131691毫秒
特征矩阵展开4分钟:40099毫秒
特征矩阵展开4平均数:54812毫秒
特征矩阵展开8分钟:40001毫秒
特征矩阵展开8平均数:51482毫秒
本征向量分钟:100270毫秒
特征向量平均值:117316毫秒
特征向量展开分钟:59966毫秒
特征向量展开平均值:65847毫秒
在每一个我测试机,exepted一个用是最老的。看起来像在新机器上小展开可能是非常有益的(结果在4倍展开时加速1.5到3.5倍,即使展开为8,16,32或256时也不会增加)。
*所有*优化应该是* per-CPU *。这一切都回到了根本问题:它太慢了吗?如果答案是*是*,那么我们就避免了过早的优化。你认为这可能是一个不成熟的优化?假设是Intel SSE还是C++ SSE? – Sebivor
它不是关于优化这些rutine,而是一般使用sse。通过每CPU的优化,你的意思是例如计数xmm寄存器和展开使用所有? – CzakCzan
您可以展开以隐藏ADDPS或FMA延迟,以及避免循环开销的前端瓶颈。例如对于一个点积:https://stackoverflow.com/questions/45113527/why-does-mulss-take-only-3-cycles-on-haswell-different-from-agners-instruction。 –