目前项目中使用的线程池(详见:http://threadpool.sourceforge...),虽然能用,但是代码复杂且很久没有人维护了。
本文结合 Thread 和 Asio,实现了一个线程池。一二十行代码,不能更简单了!
头文件:
#include <functional>
#include <iostream>
#include <mutex>
#include <thread>
#include <vector>
#define BOOST_ASIO_NO_DEPRECATED
#include "boost/asio.hpp"
线程池类:
class ThreadPool {
public:
explicit ThreadPool(std::size_t size)
: work_guard_(boost::asio::make_work_guard(io_context_)) {
workers_.reserve(size);
for (std::size_t i = 0; i < size; ++i) {
workers_.emplace_back(&boost::asio::io_context::run, &io_context_);
}
}
~ThreadPool() {
io_context_.stop();
for (auto& w : workers_) {
w.join();
}
}
// Add new work item to the pool.
template<class F>
void Enqueue(F f) {
boost::asio::post(io_context_, f);
}
private:
std::vector<std::thread> workers_;
boost::asio::io_context io_context_;
typedef boost::asio::io_context::executor_type ExecutorType;
boost::asio::executor_work_guard<ExecutorType> work_guard_;
};
成员变量 work_guard_
的作用是,让 io_context
即使在没有异步任务可执行时也保持运行(即 io_context::run
不返回)。详见 StackOverflow 的讨论:Why should I use io_service::work?
示例:
// For output.
std::mutex g_io_mutex;
int main() {
// Create a thread pool of 4 worker threads.
ThreadPool pool(4);
// Queue a bunch of work items.
for (int i = 0; i < 8; ++i) {
pool.Enqueue([i] {
{
std::lock_guard<std::mutex> lock(g_io_mutex);
std::cout << "Hello" << "(" << i << ") " << std::endl;
}
std::this_thread::sleep_for(std::chrono::seconds(1));
{
std::lock_guard<std::mutex> lock(g_io_mutex);
std::cout << "World" << "(" << i << ")" << std::endl;
}
});
}
return 0;
}
输出(每次都不一样):
Hello(0)
Hello(1)
Hello(2)
Hello(3)
<Wait about 1 second>
World(3)
World(2)
World(1)
World(0)
参考:
**粗体** _斜体_ [链接](http://example.com) `代码` - 列表 > 引用
。你还可以使用@
来通知其他用户。