Question

我有2个进程（生产者和消费者）在共享内存中共享一个int deque，我让生产者进程在deque中放入2个数字，然后它进入等待状态，失去其互斥锁。然后我让消费者进程删除数字并打印它们。然后它会对生产者正在等待的条件进行通知。消费者然后在第二个条件下自行等待。在这种情况下，生产者不会醒来。我在进程之间使用相同的互斥锁。请在下面找到所有代码。

包含文件shared_memory.h：

#ifndef SharedMemory_h
#define SharedMemory_h

#include <boost/interprocess/managed_shared_memory.hpp>
#include <boost/interprocess/containers/deque.hpp>
#include <boost/interprocess/allocators/allocator.hpp>
#include <boost/interprocess/sync/scoped_lock.hpp>
#include <boost/interprocess/offset_ptr.hpp>
#include <boost/interprocess/sync/named_condition.hpp>

using namespace boost::interprocess;

typedef allocator<offset_ptr<int>, managed_shared_memory::segment_manager> ShmemAllocator;
typedef deque<offset_ptr<int>, ShmemAllocator> Deque;

#endif

制片人流程：

#include "shared_memory.h"

struct shm_remove
{
    shm_remove() { shared_memory_object::remove("MySharedMemory"); }
    ~shm_remove() { shared_memory_object::remove("MySharedMemory"); }
} remover;

struct mutex_remove
{
    mutex_remove() { named_mutex::remove("MyMutex"); }
    ~mutex_remove() { named_mutex::remove("MyMutex"); }
} mutex_remover;

//Create shared memory, mutex and condtion
managed_shared_memory segment(create_only, "MySharedMemory", 10000000);
named_mutex mutex(create_only,"MyMutex");
named_condition full(open_or_create,"FullCondition");
named_condition empty(open_or_create,"EmptyCondition");

const ShmemAllocator alloc_inst (segment.get_segment_manager());


int main() 
{
    Deque* MyDeque;
    offset_ptr<int> a, b;
    try{
        MyDeque = segment.construct<Deque>("MyDeque")(alloc_inst);
        try{
           a = static_cast<int*> (segment.allocate(sizeof(int)));
           b = static_cast<int*> (segment.allocate(sizeof(int)));
        }catch(bad_alloc &ex){
             std::cout << "Could not allocate int" << std::endl;
        }
    }catch(bad_alloc &ex){
        std::cout << "Could not allocate queue" << std::endl;
    }
    scoped_lock<named_mutex> lock(mutex);
    while(1)
    {
        while (MyDeque->size() == 2)
        {
            full.wait(lock);
            std::cout << "unlocked producer" << std::endl;
        }

        if (MyDeque->size() == 0)
        {
            *a = 2;
            MyDeque->push_back(a);
        }

        if (MyDeque->size() == 1)
        {
            *b = 4;
            MyDeque->push_back(b);
            empty.notify_one();
        }
    }
}

消费者流程：

#include "shared_memory.h"

managed_shared_memory segment(open_only, "MySharedMemory");
Deque* MyDeque = segment.find<Deque>("MyDeque").first;

named_mutex mutex(open_only, "MyMutex");
named_condition full(open_only, "FullCondition");
named_condition empty(open_only, "EmptyCondition");

int main()
{
    scoped_lock<named_mutex> lock(mutex);
    while(1)
    {

         //volatile int size = MyDeque->size();
         while (MyDeque->size() == 0)
         {
             empty.wait(lock);
         }

         if (MyDeque->size() == 2)
         {
             std::cout << "Consumer: " << *MyDeque->front() << std::endl;
             MyDeque->pop_front();
         }

         if (MyDeque->size() == 1)
         {
             std::cout << "Consumer: " << *MyDeque->front() << std::endl;
             MyDeque->pop_front();
             full.notify_one();
         }
    }
}

虽然第一次迭代时调试事情似乎没问题，但是生产者在deque上放置数字2和4然后等待完整条件。然后消费者获得锁定，打印这些数字，在完整条件下执行notify_one然后进入等待状态。在此之后，制片人不会醒来。

Answer 1

在通知时不得锁定互斥锁。这就是僵局的原因。

boost named_condition不会唤醒等待进程

1 个答案: