我编写了代码来测试openmp on win（Win7 x64，Corei7 3.4HGz）和Mac（10.12.3 Core i7 2.7 HGz）的性能。
如果我在单个线程中运行应用程序，则时间差对应于处理器的频率比（大约）。但是如果你运行4个线程，差异变得明显：win程序比mac程序快70倍。
#include <cmath>
#include <mutex>
#include <cstdint>
#include <cstdio>
#include <iostream>
#include <omp.h>
#include <boost/chrono/chrono.hpp>
static double ActionWithNumber(double number)
{
double sum = 0.0f;
for (std::uint32_t i = 0; i < 50; i++)
{
double coeff = sqrt(pow(std::abs(number), 0.1));
double res = number*(1.0-coeff)*number*(1.0-coeff) * 3.0;
sum += sqrt(res);
}
return sum;
}
static double TestOpenMP(void)
{
const std::uint32_t len = 4000000;
double *a;
double *b;
double *c;
double sum = 0.0;
std::mutex _mutex;
a = new double[len];
b = new double[len];
c = new double[len];
for (std::uint32_t i = 0; i < len; i++)
{
c[i] = 0.0;
a[i] = sin((double)i);
b[i] = cos((double)i);
}
boost::chrono::time_point<boost::chrono::system_clock> start, end;
start = boost::chrono::system_clock::now();
double k = 2.0;
omp_set_num_threads(4);
#pragma omp parallel for
for (int i = 0; i < len; i++)
{
c[i] = k*a[i] + b[i] + k;
if (c[i] > 0.0)
{
c[i] += ActionWithNumber(c[i]);
}
else
{
c[i] -= ActionWithNumber(c[i]);
}
std::lock_guard<std::mutex> scoped(_mutex);
sum += c[i];
}
end = boost::chrono::system_clock::now();
boost::chrono::duration<double> elapsed_time = end - start;
double sum2 = 0.0;
for (std::uint32_t i = 0; i < len; i++)
{
sum2 += c[i];
c[i] /= sum2;
}
if (std::abs(sum - sum2) > 0.01) printf("Incorrect result.\n");
delete[] a;
delete[] b;
delete[] c;
return elapsed_time.count();
}
int main()
{
double sum = 0.0;
const std::uint32_t steps = 5;
for (std::uint32_t i = 0; i < steps; i++)
{
sum += TestOpenMP();
}
sum /= (double)steps;
std::cout << "Elapsed time = " << sum;
return 0;
}
我在这里专门使用互斥来比较openmp在“mac”和“win”上的表现。在“Win”函数返回0.39秒的时间。在“Mac”功能上返回25秒的时间，即慢70倍。
造成这种差异的原因是什么？
首先，感谢编辑我的帖子（我用翻译来写文字）。 在真实应用程序中，我以随机顺序更新巨大矩阵（20000х20000）中的值。每个线程确定新值并将其写入特定单元格。我为每一行创建了一个互斥锁，因为在大多数情况下，不同的线程会写入不同的行。但显然在2个线程写入一行并且存在长锁的情况下。目前我无法在不同的线程中划分行，因为记录的顺序由FEM元素决定。 因此，只需在其中放置一个关键部分，因为它将阻止对整个矩阵的写入。
我在实际应用程序中编写了代码。
static double ActionWithNumber(double number)
{
const unsigned int steps = 5000;
double sum = 0.0f;
for (u32 i = 0; i < steps; i++)
{
double coeff = sqrt(pow(abs(number), 0.1));
double res = number*(1.0-coeff)*number*(1.0-coeff) * 3.0;
sum += sqrt(res);
}
sum /= (double)steps;
return sum;
}
static double RealAppTest(void)
{
const unsigned int elementsNum = 10000;
double* matrix;
unsigned int* elements;
boost::mutex* mutexes;
elements = new unsigned int[elementsNum*3];
matrix = new double[elementsNum*elementsNum];
mutexes = new boost::mutex[elementsNum];
for (unsigned int i = 0; i < elementsNum; i++)
for (unsigned int j = 0; j < elementsNum; j++)
matrix[i*elementsNum + j] = (double)(rand() % 100);
for (unsigned int i = 0; i < elementsNum; i++) //build FEM element like Triangle
{
elements[3*i] = rand()%(elementsNum-1);
elements[3*i+1] = rand()%(elementsNum-1);
elements[3*i+2] = rand()%(elementsNum-1);
}
boost::chrono::time_point<boost::chrono::system_clock> start, end;
start = boost::chrono::system_clock::now();
omp_set_num_threads(4);
#pragma omp parallel for
for (int i = 0; i < elementsNum; i++)
{
unsigned int* elems = &elements[3*i];
for (unsigned int j = 0; j < 3; j++)
{
//in here set mutex for row with index = elems[j];
boost::lock_guard<boost::mutex> lockup(mutexes[i]);
double res = 0.0;
for (unsigned int k = 0; k < 3; k++)
{
res += ActionWithNumber(matrix[elems[j]*elementsNum + elems[k]]);
}
for (unsigned int k = 0; k < 3; k++)
{
matrix[elems[j]*elementsNum + elems[k]] = res;
}
}
}
end = boost::chrono::system_clock::now();
boost::chrono::duration<double> elapsed_time = end - start;
delete[] elements;
delete[] matrix;
delete[] mutexes;
return elapsed_time.count();
}
int main()
{
double sum = 0.0;
const u32 steps = 5;
for (u32 i = 0; i < steps; i++)
{
sum += RealAppTest();
}
sum /= (double)steps;
std::cout<<"Elapsed time = " << sum;
return 0;
}
答案 0 :(得分：3)
您正在组合两组不同的线程/同步原语 - OpenMP，它内置于编译器中并具有运行时系统，并使用std :: mutex手动创建posix互斥锁。对某些编译器/操作系统组合存在一些互操作性问题可能就不足为奇了。
我的猜测是，在缓慢的情况下，OpenMP运行时过度，以确保更高级别的正在进行的OpenMP线程任务与手动互斥锁之间没有交互，并且在紧密循环中这样做会导致戏剧性减速。
对于OpenMP框架中的类似互斥体的行为，我们可以使用关键部分：
#pragma omp parallel for
for (int i = 0; i < len; i++)
{
//...
// replacing this: std::lock_guard<std::mutex> scoped(_mutex);
#pragma omp critical
sum += c[i];
}
或显式锁：
omp_lock_t sumlock;
omp_init_lock(&sumlock);
#pragma omp parallel for
for (int i = 0; i < len; i++)
{
//...
// replacing this: std::lock_guard<std::mutex> scoped(_mutex);
omp_set_lock(&sumlock);
sum += c[i];
omp_unset_lock(&sumlock);
}
omp_destroy_lock(&sumlock);
我们得到更合理的时间：
$ time ./openmp-original
real 1m41.119s
user 1m15.961s
sys 1m53.919s
$ time ./openmp-critical
real 0m16.470s
user 1m2.313s
sys 0m0.599s
$ time ./openmp-locks
real 0m15.819s
user 1m0.820s
sys 0m0.276s
更新：使用openmp锁数组与互斥锁完全相同没有问题：
omp_lock_t sumlocks[elementsNum];
for (unsigned idx=0; idx<elementsNum; idx++)
omp_init_lock(&(sumlocks[idx]));
//...
#pragma omp parallel for
for (int i = 0; i < elementsNum; i++)
{
unsigned int* elems = &elements[3*i];
for (unsigned int j = 0; j < 3; j++)
{
//in here set mutex for row with index = elems[j];
double res = 0.0;
for (unsigned int k = 0; k < 3; k++)
{
res += ActionWithNumber(matrix[elems[j]*elementsNum + elems[k]]);
}
omp_set_lock(&(sumlocks[i]));
for (unsigned int k = 0; k < 3; k++)
{
matrix[elems[j]*elementsNum + elems[k]] = res;
}
omp_unset_lock(&(sumlocks[i]));
}
}
for (unsigned idx=0; idx<elementsNum; idx++)
omp_destroy_lock(&(sumlocks[idx]));