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58.C++多线程_线程池

约 603 字大约 2 分钟

c++ 标准并没有提供现成的线程池,只能自己封装

简单线程池(C++11 std::thread + std::queue)

#include <iostream>
#include <vector>
#include <queue>
#include <thread>
#include <mutex>
#include <condition_variable>
#include <functional>

class ThreadPool {
public:
    ThreadPool(size_t numThreads);
    ~ThreadPool();

    void enqueue(std::function<void()> task);

private:
    std::vector<std::thread> workers;  // 线程池
    std::queue<std::function<void()>> tasks;  // 任务队列

    std::mutex queueMutex;
    std::condition_variable condition;
    bool stop = false;

    void workerThread(); // 线程工作函数
};

// 线程池构造函数
ThreadPool::ThreadPool(size_t numThreads) {
    for (size_t i = 0; i < numThreads; ++i) {
        workers.emplace_back([this] { workerThread(); });
    }
}

// 线程池析构函数
ThreadPool::~ThreadPool() {
    {
        std::unique_lock<std::mutex> lock(queueMutex);
        stop = true;
    }
    condition.notify_all(); // 唤醒所有线程
    for (std::thread &worker : workers) {
        worker.join(); // 等待线程结束
    }
}

// 向任务队列添加任务
void ThreadPool::enqueue(std::function<void()> task) {
    {
        std::unique_lock<std::mutex> lock(queueMutex);
        tasks.push(std::move(task));
    }
    condition.notify_one(); // 唤醒一个线程
}

// 线程工作函数
void ThreadPool::workerThread() {
    while (true) {
        std::function<void()> task;
        {
            std::unique_lock<std::mutex> lock(queueMutex);
            condition.wait(lock, [this] { return stop || !tasks.empty(); });
            if (stop && tasks.empty()) return;
            task = std::move(tasks.front());
            tasks.pop();
        }
        task(); // 执行任务
    }
}

int main() {
    ThreadPool pool(4); // 创建 4 个线程的线程池

    for (int i = 1; i <= 10; ++i) {
        pool.enqueue([i] {
            std::cout << "任务 " << i << " 由线程 " << std::this_thread::get_id() << " 执行\n";
            std::this_thread::sleep_for(std::chrono::milliseconds(500)); // 模拟任务
        });
    }

    std::this_thread::sleep_for(std::chrono::seconds(3)); // 主线程等待任务完成
    return 0;
}

改进版线程池(std::future 支持异步返回值)

#include <iostream>
#include <vector>
#include <queue>
#include <thread>
#include <mutex>
#include <condition_variable>
#include <future>

class ThreadPool {
public:
    ThreadPool(size_t numThreads);
    ~ThreadPool();

    template <class F, class... Args>
    auto enqueue(F&& f, Args&&... args) -> std::future<typename std::invoke_result<F, Args...>::type>;

private:
    std::vector<std::thread> workers;
    std::queue<std::function<void()>> tasks;

    std::mutex queueMutex;
    std::condition_variable condition;
    bool stop = false;

    void workerThread();
};

// 构造函数
ThreadPool::ThreadPool(size_t numThreads) {
    for (size_t i = 0; i < numThreads; ++i) {
        workers.emplace_back([this] { workerThread(); });
    }
}

// 析构函数
ThreadPool::~ThreadPool() {
    {
        std::unique_lock<std::mutex> lock(queueMutex);
        stop = true;
    }
    condition.notify_all();
    for (std::thread &worker : workers) {
        worker.join();
    }
}

// 任务入队并返回 future
template <class F, class... Args>
auto ThreadPool::enqueue(F&& f, Args&&... args) -> std::future<typename std::invoke_result<F, Args...>::type> {
    using return_type = typename std::invoke_result<F, Args...>::type;

    auto task = std::make_shared<std::packaged_task<return_type()>>(
        std::bind(std::forward<F>(f), std::forward<Args>(args)...)
    );

    std::future<return_type> res = task->get_future();
    {
        std::unique_lock<std::mutex> lock(queueMutex);
        tasks.emplace([task]() { (*task)(); });
    }
    condition.notify_one();
    return res;
}

// 线程工作函数
void ThreadPool::workerThread() {
    while (true) {
        std::function<void()> task;
        {
            std::unique_lock<std::mutex> lock(queueMutex);
            condition.wait(lock, [this] { return stop || !tasks.empty(); });
            if (stop && tasks.empty()) return;
            task = std::move(tasks.front());
            tasks.pop();
        }
        task();
    }
}

int main() {
    ThreadPool pool(4);

    auto future1 = pool.enqueue([](int x) { return x * 2; }, 5);
    auto future2 = pool.enqueue([](int x, int y) { return x + y; }, 10, 20);

    std::cout << "异步任务1结果: " << future1.get() << std::endl;
    std::cout << "异步任务2结果: " << future2.get() << std::endl;

    return 0;
}