C 11 中的线程池
问题: 重复创建和删除线程的成本很高。我们如何建立一个持久的线程池来处理任务而不产生这种开销?
解决方案:
实现 ThreadPool 类
为了创建一个高效的线程池,我们首先定义ThreadPool类:
class ThreadPool { public: void Start(); void QueueJob(const std::function<void()>& job); void Stop(); bool busy(); private: void ThreadLoop(); bool should_terminate = false; std::mutex queue_mutex; std::condition_variable mutex_condition; std::vector<std::thread> threads; std::queue<std::function<void()>> jobs; };
1。 ThreadPool::Start:
根据系统能力创建固定数量的线程:
void ThreadPool::Start() { const uint32_t num_threads = std::thread::hardware_concurrency(); for (uint32_t ii = 0; ii < num_threads; ++ii) { threads.emplace_back(std::thread(&ThreadPool::ThreadLoop,this)) } }
2。 ThreadPool::ThreadLoop:
等待新任务的无限循环:
void ThreadPool::ThreadLoop() { while (true) { std::function<void()> job; { std::unique_lock<std::mutex> lock(queue_mutex); mutex_condition.wait(lock, [this] { return !jobs.empty() || should_terminate; }); if (should_terminate) { return; } job = jobs.front(); jobs.pop(); } job(); } }
3. ThreadPool::QueueJob:
向池中添加新任务:
void ThreadPool::QueueJob(const std::function<void()>& job) { { std::unique_lock<std::mutex> lock(queue_mutex); jobs.push(job); } mutex_condition.notify_one(); }
4. ThreadPool::busy:
检查池中是否有活动作业:
bool ThreadPool::busy() { bool poolbusy; { std::unique_lock<std::mutex> lock(queue_mutex); poolbusy = !jobs.empty(); } return poolbusy; }
5. ThreadPool::Stop:
优雅地停止线程池:
void ThreadPool::Stop() { { std::unique_lock<std::mutex> lock(queue_mutex); should_terminate = true; } mutex_condition.notify_all(); for (std::thread& active_thread : threads) { active_thread.join(); } threads.clear(); }
用法:
thread_pool->QueueJob([] { /* ... */ });
此实现提供了一个动态的线程池,线程持续运行并等待添加任务。
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