C++ OpenMP:循环通过';标准::地图';基准(动态调度)
我必须循环执行C++ OpenMP:循环通过';标准::地图';基准(动态调度),c++,multithreading,parallel-processing,multiprocessing,openmp,C++,Multithreading,Parallel Processing,Multiprocessing,Openmp,我必须循环执行std::map,每次迭代中必须完成的工作具有以下属性: 在每次迭代中,工作量是不同的 线程之间不需要任何同步 看起来是动态调度的完美场景,不是吗 然而,就OpenMP的循环并行化而言,非随机访问迭代器(如std::maphas)是臭名昭著的。对我来说,这段特定代码的性能将是至关重要的,因此为了寻找最有效的解决方案,我创建了以下基准: #include <omp.h> #include <iostream> #include <map> #in
std::map
,每次迭代中必须完成的工作具有以下属性:
std::map
has)是臭名昭著的。对我来说,这段特定代码的性能将是至关重要的,因此为了寻找最有效的解决方案,我创建了以下基准:
#include <omp.h>
#include <iostream>
#include <map>
#include <vector>
#define COUNT 0x00006FFF
#define UNUSED(variable) (void)(variable)
using std::map;
using std::vector;
void test1(map<int, vector<int> >& m) {
double time = omp_get_wtime();
map<int, vector<int> >::iterator iterator = m.begin();
#pragma omp parallel
#pragma omp for schedule(dynamic, 1) nowait
for (size_t i = 0; i < m.size(); ++i) {
vector<int>* v;
#pragma omp critical
v = &iterator->second;
for (size_t j = 0; j < v->size(); ++j) {
(*v)[j] = j;
}
#pragma omp critical
iterator++;
}
printf("Test #1: %f s\n", (omp_get_wtime() - time));
}
void test2(map<int, vector<int> >& m) {
double time = omp_get_wtime();
#pragma omp parallel
{
for (map<int, vector<int> >::iterator i = m.begin(); i != m.end(); ++i) {
#pragma omp single nowait
{
vector<int>& v = i->second;
for (size_t j = 0; j < v.size(); ++j) {
v[j] = j;
}
}
}
}
printf("Test #2: %f s\n", (omp_get_wtime() - time));
}
void test3(map<int, vector<int> >& m) {
double time = omp_get_wtime();
#pragma omp parallel
{
int thread_count = omp_get_num_threads();
int thread_num = omp_get_thread_num();
size_t chunk_size = m.size() / thread_count;
map<int, vector<int> >::iterator begin = m.begin();
std::advance(begin, thread_num * chunk_size);
map<int, vector<int> >::iterator end = begin;
if (thread_num == thread_count - 1)
end = m.end();
else
std::advance(end, chunk_size);
for (map<int, vector<int> >::iterator i = begin; i != end; ++i) {
vector<int>& v = i->second;
for (size_t j = 0; j < v.size(); ++j) {
v[j] = j;
}
}
}
printf("Test #3: %f s\n", (omp_get_wtime() - time));
}
int main(int argc, char** argv) {
UNUSED(argc);
UNUSED(argv);
map<int, vector<int> > m;
for (int i = 0; i < COUNT; ++i) {
m[i] = vector<int>(i);
}
test1(m);
test2(m);
test3(m);
}
我之所以发布这个问题,是因为我发现这些结果很奇怪,而且完全出乎意料:
schedule(static,1)
,即线程使用thread\u count
步长处理迭代,而不是处理大量连续迭代。代码如下:
void test4(map<int, vector<int> >& m) {
double time = omp_get_wtime();
#pragma omp parallel
{
int thread_count = omp_get_num_threads();
int thread_num = omp_get_thread_num();
size_t map_size = m.size();
map<int, vector<int> >::iterator it = m.begin();
std::advance(it, thread_num);
for (int i = thread_num; i < map_size; i+=thread_count) {
vector<int>& v = it->second;
for (size_t j = 0; j < v.size(); ++j) {
v[j] = j;
}
if( i+thread_count < map_size ) std::advance(it, thread_count);
}
}
printf("Test #4: %f s\n", (omp_get_wtime() - time));
}
在循环中,线程决定其本地迭代器在映射上的前进量。线程首先以原子方式递增计数器并获取其上一个值,从而获得迭代索引,然后根据新索引和上一个索引之间的差值使迭代器前进。循环重复,直到计数器增加到地图大小以上。您是否尝试过使用无序地图进行基准测试?@Leeor,我为什么要这样做?在这种情况下,它不会改变任何东西。这个问题与数据结构无关。它是关于并行化策略的。最后,我对这个特定任务的
无序图不感兴趣。
void test4(map<int, vector<int> >& m) {
double time = omp_get_wtime();
#pragma omp parallel
{
int thread_count = omp_get_num_threads();
int thread_num = omp_get_thread_num();
size_t map_size = m.size();
map<int, vector<int> >::iterator it = m.begin();
std::advance(it, thread_num);
for (int i = thread_num; i < map_size; i+=thread_count) {
vector<int>& v = it->second;
for (size_t j = 0; j < v.size(); ++j) {
v[j] = j;
}
if( i+thread_count < map_size ) std::advance(it, thread_count);
}
}
printf("Test #4: %f s\n", (omp_get_wtime() - time));
}
void test5(map<int, vector<int> >& m) {
double time = omp_get_wtime();
int count = 0;
#pragma omp parallel shared(count)
{
int i;
int i_old = 0;
size_t map_size = m.size();
map<int, vector<int> >::iterator it = m.begin();
#pragma omp atomic capture
i = count++;
while (i < map_size) {
std::advance(it, i-i_old);
vector<int>& v = it->second;
for (size_t j = 0; j < v.size(); ++j) {
v[j] = j;
}
i_old = i;
#pragma omp atomic capture
i = count++;
}
}
printf("Test #5: %f s\n", (omp_get_wtime() - time));
}