c++ - 使用 pthread_create 时出现 "Segmentation fault (core dumped)"

Question

所以我遇到了一个问题:当我尝试创建最后一个线程时，它总是说核心已转储。如果我编写创建 5 个或 2 个线程，这并不重要。这是我的代码:UPD:现在我不能执行超过 3 个线程，并且线程不执行我希望它们执行的功能(消费和生产)

UPD_2:现在我收到了这样的消息:在抛出“递归调用的终止”实例后终止调用终止递归调用中止(核心转储)

 #include<cstdlib>
#include <iostream>
#include <string>
#include <mutex>
#include <pthread.h>
#include <condition_variable>

#define NUM_THREADS 4

using namespace std;

struct thread_data
{
   int  thread_id;
   int  repeat;
};


 class our_monitor{
    private:
        int buffer[100];
        mutex m;
        int n = 0, lo = 0, hi = 0;
        condition_variable in,out;
        unique_lock<mutex> lk;

    public:
        our_monitor():lk(m)
        {

        }
        void insert(int val, int repeat)
        {
            in.wait(lk, [&]{return n <= 100-repeat;});
            for(int i=0; i<repeat; i++)
            {
                buffer[hi] = val;
                hi = (hi + 1) % 100;        //ring buffer
                n = n +1;         //one more item in buffer
            }
            lk.unlock();
            out.notify_one();
        }

        int remove(int repeat)
        {
            out.wait(lk, [&]{return n >= repeat;});
            int val;
            for(int i=0; i<repeat; i++)
            {
                val = buffer[lo];
                lo = (lo + 1) % 100;
                n -= 1;
            }
            lk.unlock();
            in.notify_one();
            return val;
        }
};

our_monitor mon;

void* produce(void *threadarg)
{
    struct thread_data *my_data;
    my_data = (struct thread_data *) threadarg;
    cout<<"IN produce after paramiters"<< my_data->repeat<<endl;
    int item;
    item = rand()%100 + 1;
    mon.insert(item, my_data->repeat);
    cout<< "Item: "<< item << " Was prodused by thread:"<< my_data->thread_id << endl;
}

void* consume(void *threadarg)
    {
        struct thread_data *my_data;
        my_data = (struct thread_data *) threadarg;
        cout<<"IN consume after paramiters"<< my_data->repeat<<endl;
        int item;
        item = mon.remove(my_data->repeat);
        if(item) cout<< "Item: "<< item << " Was consumed by thread:"<< my_data->thread_id << endl;
    }

int main()
{
    our_monitor *mon = new our_monitor();
    pthread_t threads[NUM_THREADS];
    thread_data td[NUM_THREADS];
    int rc;
    int i;

    for( i = 0; i < NUM_THREADS; i++ )
    {
        td[i].thread_id = i;
        td[i].repeat = rand()%5 + 1;
        if(i % 2 == 0)
        {
            cout << "main() : creating produce thread, " << i << endl;
            rc = pthread_create(&threads[i], NULL, produce, (void*) &td[i]);

            if (rc)
            {
                cout << "Error:unable to create thread," << rc << endl;
                exit(-1);
            }
        } else
        {
            cout << "main() : creating consume thread, " << i << endl;
            rc = pthread_create(&threads[i], NULL, consume, (void *)&td[i]);

            if (rc)
            {
                cout << "Error:unable to create thread," << rc << endl;
                exit(-1);
            }
        }
    }
    pthread_join(threads[0], NULL);
    pthread_join(threads[1], NULL);
    pthread_join(threads[2], NULL);
    //pthread_exit(NULL);

}

UPD:现在我不能执行超过 3 个线程，并且线程不执行我希望它们执行的功能(消费和生产)

UPD_2:现在我收到了这样的消息:在抛出“递归调用的终止”实例后终止调用终止递归调用中止(核心转储)

Answer 1

来自 cppref 关于 std::condition_variable.wait(...)

"Calling this function if lock.mutex() is not locked by the current thread is undefined behavior."

http://en.cppreference.com/w/cpp/thread/condition_variable/wait

不幸的是，程序并没有在第 47 行崩溃，而是在第 55 行崩溃，此时您解锁了未锁定的锁。

进入功能时锁定锁。我已经快速检查了你的其余逻辑，我 85% 确定它在其他方面没问题。

虽然你在这里，但这并不是绝对必要的，但这是一个很好的做法。 std::lock_guard 和 std::unique_lock 在互斥锁进入作用域时自动锁定互斥锁，并在其离开作用域时将其解锁。这有助于简化异常处理和奇怪的函数返回。我建议您摆脱 lk 作为成员变量，并将其用作作用域局部变量。

    void insert(int val, int repeat)
    {
        { // Scoped. Somewhat pedantic in this case, but it's always best to signal after the mutex is unlocked
            std::unique_lock<std::mutex> lk(m);
            in.wait(lk, [&]{return n <= 100-repeat;});
            for(int i=0; i<repeat; i++)
            {
                buffer[hi] = val;
                hi = (hi + 1) % 100;        //ring buffer
                n = n +1;         //one more item in buffer
            }
        }
        out.notify_one();
    }

好的，现在进入最后一期。生产者/消费者的最酷之处在于我们可以同时生产和消费。然而，我们只是锁定了我们的功能，所以这不再可能。您现在可以做的是将条件锁定/等待/解锁/工作/信号内部移动到for循环

伪代码:

// produce:
while (true)
{
    {
        unique_lock lk(m)
        wait(m, predicate)
    }
    produce 1
    signal
}

这相当于使用信号量(C++'11 STL 没有，但您可以轻松地创建自己的信号量，如上所示。)

// produce:
semaphore in(100);
semaphore out(0);
while (true)
{
    in.down(1)  // Subtracts 1 from in.count. Blocks when in.count == 0 (meaning the buffer is full)
    produce 1
    out.up(1)   // Adds 1 to out.count
}