尝试实现一个简单的threadpool

参考资料：
C++高并发多线程学习(一)
C++ 并发编程[Part 2]

---

头文件：

#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 

using namespace std;

class Task
{
public:
    Task(){}
    ~Task(){}

    virtual void run() = 0;    // The specific work is realized by the developer
};

//从Task类继承，实现run()
class MyTask: public Task
{
public:
    MyTask(int num) {number = num; }
    ~MyTask();

    virtual void run(){ cout<<"task output: "<<number<<endl; }

private:
    int number;
};


class threadpool
{
public:
    threadpool(int threadCount=4);
    ~threadpool();

    void    addTask(Task* pTask);

private:
    atomic<bool>    m_bThreadRun;
    condition_variable m_ThreadCondition;
    mutex           m_Mutex;
    queue<Task*>    m_TaskQue;
    vector<thread>  m_ThreadList;

    void    run();
};

源文件：

#include "threadpool.h"

threadpool::threadpool(int threadCount)
{
    m_ThreadList.reserve(threadCount);
    for(int i=0; i<threadCount; i++)
    {
        m_ThreadList.emplace_back(&threadpool::run, this); //这种构造方式是函数加参数吗
//        thread t(&threadpool::run, this);
//        m_ThreadList.push_back(t);         //为什么通过这种方式构造编译报错？
    }
}

threadpool::~threadpool()
{
    m_bThreadRun.store(false);
    m_ThreadCondition.notify_all();

    for(int i=0; i<m_ThreadList.size(); i++)
    {
        m_ThreadList.at(i).join();
    }

    while(!m_TaskQue.empty())
    {
        Task* pTask = m_TaskQue.front();
        m_TaskQue.pop();

        if(pTask)
        {
            delete pTask;
        }
    }
}

void threadpool::addTask(Task* pTask)
{
    if(pTask)
    {
        unique_lock<mutex> lock(m_Mutex);
        m_TaskQue.push(pTask);
    }
    else
        return;
    m_ThreadCondition.notify_one();
}

void threadpool::run()
{
    m_bThreadRun.store(true);
    while(m_bThreadRun.load())
    {
        unique_lock<mutex> lock(m_Mutex);
        //任务队列不为空时直接返回
        m_ThreadCondition.wait(lock, [this]{return !m_TaskQue.empty() || !m_bThreadRun.load(); });

        if(m_TaskQue.empty())
            continue;
        Task* pTask = m_TaskQue.front();
        m_TaskQue.pop();
        lock.unlock(); //不解锁的话，是在task的run结束后才解锁任务队列，实际上只有一个线程在执行任务
        if(pTask)
        {
            pTask->run();
            delete pTask;
            pTask = nullptr;
        }

        this_thread::sleep_for(chrono::milliseconds(2));
    }
}

测试：

threadpool pool(4);

for(int i=0; i<1000; i++)
{
    Task* pTask = new MyTask(i);
    pool.addTask(pTask);
}

希望给予指正