C++ C++;带boost-asio的线程
我有一个使用boosts库的线程池,我在下面的示例中设置了两个线程,它们可以重新运行4次。我可以检查C++ C++;带boost-asio的线程,c++,multithreading,boost,C++,Multithreading,Boost,我有一个使用boosts库的线程池,我在下面的示例中设置了两个线程,它们可以重新运行4次。我可以检查的最佳方法是_service,查看是否所有子线程都已执行完毕,然后才能继续编写代码?代码的其余部分取决于在程序继续之前完成的所有子线程。如果我进行Sleep(1000)调用,我可以获得所需的行为,但这是不可取的,我查看了io\u service\uu.stopped()上的检查,但总是返回0。任何想法都将不胜感激 #include <iostream>
的最佳方法是_service
,查看是否所有子线程都已执行完毕,然后才能继续编写代码?代码的其余部分取决于在程序继续之前完成的所有子线程。如果我进行Sleep(1000)
调用,我可以获得所需的行为,但这是不可取的,我查看了io\u service\uu.stopped()
上的检查,但总是返回0
。任何想法都将不胜感激
#include <iostream>
#include <boost/asio/io_service.hpp>
#include <boost/bind.hpp>
#include <boost/thread/thread.hpp>
#include <vector>
#include <string>
using namespace std;
class Model {
public:
// Constructor
Model() {
work_ctrl_ = new boost::asio::io_service::work(io_service_);
for (int i = 0; i < 2; ++i) {
threads_.create_thread(
boost::bind(&boost::asio::io_service::run, &io_service_));
}
}
// Deconstructor
~Model() {
delete work_ctrl_;
}
// Function I want to thread
void manipulate_vector(unsigned start, unsigned last) {
cout << "entering manipulate vector(), from thread " << boost::this_thread::get_id() << endl;
for(unsigned k = start; k <= last; ++k)
my_vector_[k] *= sqrt(32);
cout << "exit manipulate vector()" << endl;
Sleep(500); // Add a sleep to mimic a long algorithm being executed
}
void update() {
// Do otherstuff that can't be threaded
cout << "entering update" << endl;
// run manipulate_vector() across multiple threads
// - start thread
// - execute function call.
// - stop thread
io_service_.post(boost::bind(manipulate_vector, this, 0, mid_point_));
io_service_.post(boost::bind(manipulate_vector, this, mid_point_, my_vector_.size()));
cout << io_service_.stopped() << endl;
// keep doing otherstuff that can't be threaded
cout << "hopefully the threads are finished and I can take that information and continue." << endl;
}
void run(void) {
// call update 10 times
for(unsigned i = 0; i < 4; ++i) {
update();
//Sleep(2000);
}
}
void initialise() {
// initialise vector
for(unsigned j = 0; j < 100000000; ++j)
my_vector_.push_back(j);
mid_point_ = 49999999;
}
private:
boost::asio::io_service io_service_;
boost::thread_group threads_;
boost::asio::io_service::work *work_ctrl_;
unsigned n_threads_;
vector<double> my_vector_;
unsigned mid_point_;
};
int main() {
std::cout << "----------Enter Main----------" << std::endl;
Model model;
model.initialise();
model.run();
std::cout << "----------Exit Main----------" << std::endl;
system("PAUSE");
}
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使用名称空间std;
类模型{
公众:
//建造师
模型(){
work\u ctrl\uu=newboost::asio::io\u服务::work(io\u服务);
对于(int i=0;i<2;++i){
线程。创建线程(
boost::绑定(&boost::asio::io_服务::运行,&io_服务);
}
}
//解构器
~Model(){
删除工作\u ctrl\u;
}
//我想要线程的函数
无效向量(无符号开始,无符号最后){
您可能想知道工作何时完成,而不是线程何时停止执行(直到您调用io\u service.stop()
或删除work-ctrl
),线程才会停止执行)
标准解决方案是使用条件变量:
std::mutex mutex;
std::condition_variable condition;
int workCount;
void manipulate_vector(unsigned start, unsigned last) {
cout << "entering manipulate vector(), from thread " << boost::this_thread::get_id() << endl;
for(unsigned k = start; k <= last; ++k)
my_vector_[k] *= sqrt(32);
cout << "exit manipulate vector()" << endl;
Sleep(500); // Add a sleep to mimic a long algorithm being executed
std::unique_lock<std::mutex> lock(mutex);
workCount--;
condition.notify_one();
}
void update() {
// Do otherstuff that can't be threaded
cout << "entering update" << endl;
// run manipulate_vector() across multiple threads
// - start thread
// - execute function call.
// - stop thread
workCount = 2;
io_service_.post(boost::bind(manipulate_vector, this, 0, mid_point_));
io_service_.post(boost::bind(manipulate_vector, this, mid_point_, my_vector_.size()));
{
std::unique_lock<std::mutex> lock(mutex);
condition.wait(lock, [&]{ return workCount == 0; });
}
}
std::mutex mutex;
std::条件\可变条件;
整数工作计数;
无效向量(无符号开始,无符号最后){
cout如果对当前线程中正在运行的某些工作没有问题,可以调用io\u service\uu.run()
删除工作\u ctrl\ucode>后要等待的位置。当它返回时,没有剩余的工作,您可以安全地继续。您可以将线程组中的并发级别降低1,以获得相同的总体并发性
一个简单的替代方法是调用threads\uu.join\uall()
在所有帖子排队后,您删除了work\u ctrl\u
我不确定您的意思,我的印象是,如果您加入线程,您就不能像我的示例中那样重新加入。@Cyrillm\u 44是的,在加入线程后,您必须创建新的线程