下面是一个小程序,用于并行化 1/(n^2) 级数的近似值。请注意全局参数 NUM_THREADS
。
我的问题是,将线程数从 1 增加到 4(我的计算机的处理器数是 4)不会显着影响计时实验的结果。您是否看到 ThreadFunction
中存在逻辑缺陷?是否存在导致序列化执行的错误共享或错位阻塞?
#include <iostream>
#include <thread>
#include <vector>
#include <mutex>
#include <string>
#include <future>
#include <chrono>
std::mutex sum_mutex; // This mutex is for the sum vector
std::vector<double> sum_vec; // This is the sum vector
int NUM_THREADS = 1;
int UPPER_BD = 1000000;
/* Thread function */
void ThreadFunction(std::vector<double> &l, int beg, int end, int thread_num)
{
double sum = 0;
for(int i = beg; i < end; i++) sum += (1 / ( l[i] * l[i]) );
std::unique_lock<std::mutex> lock1 (sum_mutex, std::defer_lock);
lock1.lock();
sum_vec.push_back(sum);
lock1.unlock();
}
void ListFill(std::vector<double> &l, int z)
{
for(int i = 0; i < z; ++i) l.push_back(i);
}
int main()
{
std::vector<double> l;
std::vector<std::thread> thread_vec;
ListFill(l, UPPER_BD);
int len = l.size();
int lower_bd = 1;
int increment = (UPPER_BD - lower_bd) / NUM_THREADS;
for (int j = 0; j < NUM_THREADS; ++j)
{
thread_vec.push_back(std::thread(ThreadFunction, std::ref(l), lower_bd, lower_bd + increment, j));
lower_bd += increment;
}
for (auto &t : thread_vec) t.join();
double big_sum;
for (double z : sum_vec) big_sum += z;
std::cout << big_sum << std::endl;
return 0;
}
最佳答案
通过查看您的代码,我怀疑 ListFill 花费的时间比 ThreadFunction 长。为什么将值列表而不是每个线程应该循环的边界传递给线程?像这样的东西:
void ThreadFunction( int beg, int end ) {
double sum = 0.0;
for(double i = beg; i < end; i++)
sum += (1.0 / ( i * i) );
std::unique_lock<std::mutex> lock1 (sum_mutex);
sum_vec.push_back(sum);
}
为了最大化并行性,您需要将尽可能多的工作推送到线程上。参见 Amdahl's Law
关于c++ - 线程未能影响性能,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/25086700/