线程池未完成所有任务
Thread pool not completing all tasks
我之前问过这个问题的简单版本并得到了正确答案:
现在,我正在尝试使用线程池从 class 的对象并行执行 运行 任务。我的任务很简单,只为 class 的实例打印一个数字。我期待数字 0->9 被打印出来,但我得到的是一些数字不止一次打印,而一些数字根本没有打印。谁能看出我在循环中创建任务时做错了什么?
#include "iostream"
#include "ThreadPool.h"
#include <chrono>
#include <thread>
using namespace std;
using namespace dynamicThreadPool;
class test {
int x;
public:
test(int x_in) : x(x_in) {}
void task()
{
cout << x << endl;
}
};
int main(void)
{
thread_pool pool;
for (int i = 0; i < 10; i++)
{
test* myTest = new test(i);
std::function<void()> myFunction = [&] {myTest->task(); };
pool.submit(myFunction);
}
while (!pool.isQueueEmpty())
{
std::this_thread::sleep_for(std::chrono::milliseconds(1000));
cout << "waiting for tasks to complete" << endl;
}
return 0;
}
这是我的线程池,我从《C++ 并发实战》一书中得到了这个定义:
#pragma once
#include <queue>
#include <future>
#include <list>
#include <functional>
#include <memory>
template<typename T>
class threadsafe_queue
{
private:
mutable std::mutex mut;
std::queue<T> data_queue;
std::condition_variable data_cond;
public:
threadsafe_queue() {}
void push(T new_value)
{
std::lock_guard<std::mutex> lk(mut);
data_queue.push(std::move(new_value));
data_cond.notify_one();
}
void wait_and_pop(T& value)
{
std::unique_lock<std::mutex> lk(mut);
data_cond.wait(lk, [this] {return !data_queue.empty(); });
value = std::move(data_queue.front());
data_queue.pop();
}
bool try_pop(T& value)
{
std::lock_guard<std::mutex> lk(mut);
if (data_queue.empty())
return false;
value = std::move(data_queue.front());
data_queue.pop();
return true;
}
bool empty() const
{
std::lock_guard<std::mutex> lk(mut);
return data_queue.empty();
}
};
class join_threads
{
std::vector<std::thread>& threads;
public:
explicit join_threads(std::vector<std::thread>& threads_) : threads(threads_) {}
~join_threads()
{
for (unsigned long i = 0; i < threads.size(); i++)
{
if (threads[i].joinable())
{
threads[i].join();
}
}
}
};
class thread_pool
{
std::atomic_bool done;
threadsafe_queue<std::function<void()> > work_queue;
std::vector<std::thread> threads;
join_threads joiner;
void worker_thread()
{
while (!done)
{
std::function<void()> task;
if (work_queue.try_pop(task))
{
task();
}
else
{
std::this_thread::yield();
}
}
}
public:
thread_pool() : done(false), joiner(threads)
{
unsigned const thread_count = std::thread::hardware_concurrency();
try
{
for (unsigned i = 0; i < thread_count; i++)
{
threads.push_back(std::thread(&thread_pool::worker_thread, this));
}
}
catch (...)
{
done = true;
throw;
}
}
~thread_pool()
{
done = true;
}
template<typename FunctionType>
void submit(FunctionType f)
{
work_queue.push(std::function<void()>(f));
}
bool isQueueEmpty()
{
return work_queue.empty();
}
};
代码太多,无法全部分析,但您在这里引用了一个指针:
{
test* myTest = new test(i);
std::function<void()> myFunction = [&] {myTest->task(); };
pool.submit(myFunction);
} // pointer goes out of scope
在该指针超出范围后,如果您稍后执行 myTest->task();
.
,您将有 未定义的行为
要解决眼前的问题,请复制指针和 delete
之后的对象,以免内存泄漏:
{
test* myTest = new test(i);
std::function<void()> myFunction = [=] {myTest->task(); delete myTest; };
pool.submit(myFunction);
}
我怀疑根本不用 new
就可以解决这个问题,但我会把它留给你。
我之前问过这个问题的简单版本并得到了正确答案:
#include "iostream"
#include "ThreadPool.h"
#include <chrono>
#include <thread>
using namespace std;
using namespace dynamicThreadPool;
class test {
int x;
public:
test(int x_in) : x(x_in) {}
void task()
{
cout << x << endl;
}
};
int main(void)
{
thread_pool pool;
for (int i = 0; i < 10; i++)
{
test* myTest = new test(i);
std::function<void()> myFunction = [&] {myTest->task(); };
pool.submit(myFunction);
}
while (!pool.isQueueEmpty())
{
std::this_thread::sleep_for(std::chrono::milliseconds(1000));
cout << "waiting for tasks to complete" << endl;
}
return 0;
}
这是我的线程池,我从《C++ 并发实战》一书中得到了这个定义:
#pragma once
#include <queue>
#include <future>
#include <list>
#include <functional>
#include <memory>
template<typename T>
class threadsafe_queue
{
private:
mutable std::mutex mut;
std::queue<T> data_queue;
std::condition_variable data_cond;
public:
threadsafe_queue() {}
void push(T new_value)
{
std::lock_guard<std::mutex> lk(mut);
data_queue.push(std::move(new_value));
data_cond.notify_one();
}
void wait_and_pop(T& value)
{
std::unique_lock<std::mutex> lk(mut);
data_cond.wait(lk, [this] {return !data_queue.empty(); });
value = std::move(data_queue.front());
data_queue.pop();
}
bool try_pop(T& value)
{
std::lock_guard<std::mutex> lk(mut);
if (data_queue.empty())
return false;
value = std::move(data_queue.front());
data_queue.pop();
return true;
}
bool empty() const
{
std::lock_guard<std::mutex> lk(mut);
return data_queue.empty();
}
};
class join_threads
{
std::vector<std::thread>& threads;
public:
explicit join_threads(std::vector<std::thread>& threads_) : threads(threads_) {}
~join_threads()
{
for (unsigned long i = 0; i < threads.size(); i++)
{
if (threads[i].joinable())
{
threads[i].join();
}
}
}
};
class thread_pool
{
std::atomic_bool done;
threadsafe_queue<std::function<void()> > work_queue;
std::vector<std::thread> threads;
join_threads joiner;
void worker_thread()
{
while (!done)
{
std::function<void()> task;
if (work_queue.try_pop(task))
{
task();
}
else
{
std::this_thread::yield();
}
}
}
public:
thread_pool() : done(false), joiner(threads)
{
unsigned const thread_count = std::thread::hardware_concurrency();
try
{
for (unsigned i = 0; i < thread_count; i++)
{
threads.push_back(std::thread(&thread_pool::worker_thread, this));
}
}
catch (...)
{
done = true;
throw;
}
}
~thread_pool()
{
done = true;
}
template<typename FunctionType>
void submit(FunctionType f)
{
work_queue.push(std::function<void()>(f));
}
bool isQueueEmpty()
{
return work_queue.empty();
}
};
代码太多,无法全部分析,但您在这里引用了一个指针:
{
test* myTest = new test(i);
std::function<void()> myFunction = [&] {myTest->task(); };
pool.submit(myFunction);
} // pointer goes out of scope
在该指针超出范围后,如果您稍后执行 myTest->task();
.
要解决眼前的问题,请复制指针和 delete
之后的对象,以免内存泄漏:
{
test* myTest = new test(i);
std::function<void()> myFunction = [=] {myTest->task(); delete myTest; };
pool.submit(myFunction);
}
我怀疑根本不用 new
就可以解决这个问题,但我会把它留给你。