C ++创建一个工作容器,您也可以添加函数,该函数将在线程中启动,完成后将其删除

问题描述 投票:0回答:1

我想创建一个容器,可以将函数推入该容器,该容器将立即在线程中启动。功能完成后,应自动将其从容器中移除,以免容器无限期地增长。

到目前为止是我的尝试:

#include <thread>
#include <future>
#include <iostream>
#include <map>

class j_thread {
    std::thread thread;
public:

    j_thread() {}

    template<typename F>
    j_thread(const F& f) : thread(f) {}

    j_thread& operator = (j_thread&& other) {
        this->thread.swap(other.thread);
        return *this;
    }

    virtual ~j_thread() {
        thread.join();
    }
};

class jobs {

    std::map<size_t, j_thread> threads;

public:

    template<typename F>
    void add_job(const F &function) {

        size_t job_id = threads.size();

        auto wrapped_function = [&function, job_id, this]() {
            function();
            threads.erase(job_id);
        };

        threads[job_id] = j_thread(wrapped_function);
    }

    void wait_for_all() {
        while(threads.size() != 0) {}
    }
};


int main() {

    jobs j;
    j.add_job([](){std::cout << "hello" << std::endl;});
    j.add_job([](){std::cout << "world" << std::endl;});
    j.wait_for_all();
}

但是运行时出现错误:

terminate called after throwing an instance of 'std::system_error'
  what():  Invalid argument
hello
terminate called recursively
12:15:44: The program has unexpectedly finished.
c++ multithreading pthreads c++17
1个回答
1
投票

在线程体内调用join是未定义的行为。

查看join的错误条件:

错误条件resource_deadlock_would_occur如果this-> get_id()==std :: this_thread :: get_id()(检测到死锁)

您的身体是:

    auto wrapped_function = [&function, job_id, this]() {
        function();
        threads.erase(job_id);
    };

在您调用erase的地方,正在调用jthread的dtor,它在可连接线程上调用了join

而不是join,在您应该呼叫detach

为避免悬空参考,必须按值捕获function

[另外,调用sizeerase时,您还必须添加一些互斥体以避免地图上的数据争用:

std::mutex m;

int size() {
    std::lock_guard<std::mutex> lock{m};
    return threads.size();
}

auto wrapped_function = [f = function, job_id, this]() {
    f();
    std::lock_guard<std::mutex> l(m);
    threads.erase(job_id);
};

void wait_for_all() {
    while(size() != 0) {}
}

Demo

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