假设有一个像这样的线程
void mythread()
{
int res;
while(1) {
{
boost::lock_guard<boost::mutex> lock(mylock);
res = do_my_stuff();
}
boost::this_thread::sleep(boost::posix_time::seconds(5));
}
}
并且线程正在休眠。如果在线程之外发生了某些事情,我希望能够增加睡眠时间。
最好的方法是什么?
答案 0 :(得分:1)
condition_variable
表示对截止日期的更改这有利于支持超时缩短
的情况。#include <thread>
#include <chrono>
#include <iostream>
#include <condition_variable>
namespace demo
{
namespace chrono = std::chrono;
using our_clock = chrono::system_clock;
struct Worker
{
mutable std::mutex _mx;
// shared, protected by _mx:
our_clock::time_point _deadline;
mutable std::condition_variable _cv;
Worker(our_clock::time_point deadline) : _deadline(deadline) {}
void operator()() const {
std::unique_lock<std::mutex> lk(_mx);
_cv.wait_until(lk, _deadline, [this]
{
std::cout << "worker: Signaled\n";
auto now = our_clock::now();
if (now >= _deadline)
return true;
std::cout << "worker: Still waiting " << chrono::duration_cast<chrono::milliseconds>(_deadline - now).count() << "ms...\n";
return false;
});
std::cout << "worker: Done\n";
}
};
}
int main()
{
using namespace demo;
Worker worker(our_clock::now() + chrono::seconds(2));
auto th = std::thread(std::cref(worker));
// after 2 seconds, update the timepoint
std::this_thread::sleep_for(chrono::seconds(1));
{
std::lock_guard<std::mutex> lk(worker._mx);
std::cout << "Updating shared delay value..." << "\n";
worker._deadline = our_clock::now() + chrono::seconds(1);
worker._cv.notify_all();
}
th.join();
}
这是一种仅限标准库的方法,在截止日期前不使用同步。
我更倾向于使用原子time_point
作为截止日期值本身,但这不受支持。接下来最好的事情是shared_ptr<time_point>
(std::atomic_load
/ atomic_store
)但我的编译器库尚未实现(grrr)。
所以,相反,我从开始时间开始共享'偏移':
#include <thread>
#include <chrono>
#include <iostream>
#include <atomic>
namespace demo
{
namespace chrono = std::chrono;
using our_clock = chrono::system_clock;
using shared_delay = std::atomic<our_clock::duration>;
void worker(our_clock::time_point const start, shared_delay const& delay)
{
for (our_clock::time_point deadline; our_clock::now() < (deadline = start + delay.load());)
{
std::cout << "worker: Sleeping for " << chrono::duration_cast<chrono::milliseconds>(deadline - our_clock::now()).count() << "ms...\n";
std::this_thread::sleep_until(deadline);
}
std::cout << "worker: Done\n";
}
}
int main()
{
using namespace demo;
shared_delay delay(chrono::seconds(2));
auto th = std::thread(worker, our_clock::now(), std::cref(delay));
// after 2 seconds, update the timepoint
std::this_thread::sleep_for(chrono::seconds(1));
std::cout << "Updating shared delay value..." << "\n";
delay.store(chrono::seconds(3));
th.join();
}
答案 1 :(得分:0)
这是一个快速而肮脏的方法:
volatile bool someCondition = false;
void callFromOtherThread(bool x) {
boost::lock_guard<boost::mutex> lock(mylock2);
someCondition = x;
}
void mythread()
{
int res;
while(1) {
bool keepwaiting = false;
{
boost::lock_guard<boost::mutex> lock(mylock2);
keepwaiting = someCondition;
}
if (!keepwaiting) {
boost::lock_guard<boost::mutex> lock(mylock);
res = do_my_stuff();
}
boost::this_thread::sleep(boost::posix_time::seconds(5));
}
}
当你的线程完成睡眠时,会检查它发生了“其他事情”,如果发生了,它会跳过'do_my_stuff()'并再次回到睡眠状态。
我怀疑有关您的用例的更多信息,可能有可能重写事物以使用条件变量。