我正在尝试在C ++ 11中实现一个无锁多个生产者，多个消费者队列。我这样做是为了学习练习，所以我很清楚我可以使用现有的开源实现，但我真的很想知道为什么我的代码不起作用。数据存储在一个环形缓冲区中，显然它是一个“有界MPMC队列”。
我已将其与Disruptor所读到的内容非常接近。我注意到的事情是它对单个消费者和单个/多个生产者来说绝对正常，它只是多个消费者似乎打破了它。
这是队列：
template <typename T>
class Queue : public IQueue<T>
{
public:
explicit Queue( int capacity );
~Queue();
bool try_push( T value );
bool try_pop( T& value );
private:
typedef struct
{
bool readable;
T value;
} Item;
std::atomic<int> m_head;
std::atomic<int> m_tail;
int m_capacity;
Item* m_items;
};
template <typename T>
Queue<T>::Queue( int capacity ) :
m_head( 0 ),
m_tail( 0 ),
m_capacity(capacity),
m_items( new Item[capacity] )
{
for( int i = 0; i < capacity; ++i )
{
m_items[i].readable = false;
}
}
template <typename T>
Queue<T>::~Queue()
{
delete[] m_items;
}
template <typename T>
bool Queue<T>::try_push( T value )
{
while( true )
{
// See that there's room
int tail = m_tail.load(std::memory_order_acquire);
int new_tail = ( tail + 1 );
int head = m_head.load(std::memory_order_acquire);
if( ( new_tail - head ) >= m_capacity )
{
return false;
}
if( m_tail.compare_exchange_weak( tail, new_tail, std::memory_order_acq_rel ) )
{
// In try_pop, m_head is incremented before the reading of the value has completed,
// so though we've acquired this slot, a consumer thread may be in the middle of reading
tail %= m_capacity;
std::atomic_thread_fence( std::memory_order_acquire );
while( m_items[tail].readable )
{
}
m_items[tail].value = value;
std::atomic_thread_fence( std::memory_order_release );
m_items[tail].readable = true;
return true;
}
}
}
template <typename T>
bool Queue<T>::try_pop( T& value )
{
while( true )
{
int head = m_head.load(std::memory_order_acquire);
int tail = m_tail.load(std::memory_order_acquire);
if( head == tail )
{
return false;
}
int new_head = ( head + 1 );
if( m_head.compare_exchange_weak( head, new_head, std::memory_order_acq_rel ) )
{
head %= m_capacity;
std::atomic_thread_fence( std::memory_order_acquire );
while( !m_items[head].readable )
{
}
value = m_items[head].value;
std::atomic_thread_fence( std::memory_order_release );
m_items[head].readable = false;
return true;
}
}
}
这是我正在使用的测试：
void Test( std::string name, Queue<int>& queue )
{
const int NUM_PRODUCERS = 64;
const int NUM_CONSUMERS = 2;
const int NUM_ITERATIONS = 512;
bool table[NUM_PRODUCERS*NUM_ITERATIONS];
memset(table, 0, NUM_PRODUCERS*NUM_ITERATIONS*sizeof(bool));
std::vector<std::thread> threads(NUM_PRODUCERS+NUM_CONSUMERS);
std::chrono::system_clock::time_point start, end;
start = std::chrono::system_clock::now();
std::atomic<int> pop_count (NUM_PRODUCERS * NUM_ITERATIONS);
std::atomic<int> push_count (0);
for( int thread_id = 0; thread_id < NUM_PRODUCERS; ++thread_id )
{
threads[thread_id] = std::thread([&queue,thread_id,&push_count]()
{
int base = thread_id * NUM_ITERATIONS;
for( int i = 0; i < NUM_ITERATIONS; ++i )
{
while( !queue.try_push( base + i ) ){};
push_count.fetch_add(1);
}
});
}
for( int thread_id = 0; thread_id < ( NUM_CONSUMERS ); ++thread_id )
{
threads[thread_id+NUM_PRODUCERS] = std::thread([&]()
{
int v;
while( pop_count.load() > 0 )
{
if( queue.try_pop( v ) )
{
if( table[v] )
{
std::cout << v << " already set" << std::endl;
}
table[v] = true;
pop_count.fetch_sub(1);
}
}
});
}
for( int i = 0; i < ( NUM_PRODUCERS + NUM_CONSUMERS ); ++i )
{
threads[i].join();
}
end = std::chrono::system_clock::now();
std::chrono::duration<double> duration = end - start;
std::cout << name << " " << duration.count() << std::endl;
std::atomic_thread_fence( std::memory_order_acq_rel );
bool result = true;
for( int i = 0; i < NUM_PRODUCERS * NUM_ITERATIONS; ++i )
{
if( !table[i] )
{
std::cout << "failed at " << i << std::endl;
result = false;
}
}
std::cout << name << " " << ( result? "success" : "fail" ) << std::endl;
}
任何朝着正确方向的推动都将非常感激。我对内存栅栏很陌生，而不仅仅是使用互斥锁，所以我可能只是从根本上误解了一些东西。
干杯 Ĵ
答案 0 :(得分：10)
我会看看Moody Camel的实施情况。
这是一个快速通用的无锁无条件队列，完全用C ++ 11编写。文档似乎相当不错，还有一些性能测试。
在所有其他有趣的事情中（无论如何它们都值得一读），它们都包含在一个标题中，并且可以在简化的BSD许可下使用。只需将其放入您的项目中即可享受！
答案 1 :(得分：0)
这是无内存排序的无锁队列，但这在初始化队列时需要预先设置当前线程的数量。
例如：-
int* ret;
int max_concurrent_thread = 16;
lfqueue_t my_queue;
lfqueue_init(&my_queue, max_concurrent_thread );
/** Wrap This scope in other threads **/
int_data = (int*) malloc(sizeof(int));
assert(int_data != NULL);
*int_data = i++;
/*Enqueue*/
while (lfqueue_enq(&my_queue, int_data) == -1) {
printf("ENQ Full ?\n");
}
/** Wrap This scope in other threads **/
/*Dequeue*/
while ( (int_data = lfqueue_deq(&my_queue)) == NULL) {
printf("DEQ EMPTY ..\n");
}
// printf("%d\n", *(int*) ret );
free(ret);
/** End **/
lfqueue_destroy(&my_queue);
答案 2 :(得分：0)
在另一个类似的问题上，我向这个问题提出了a solution。我相信它是迄今为止发现的最小的。
我在这里不会给出相同的答案，但是the repository具有您想要的无锁队列的功能齐全的C ++实现。
编辑：感谢@PeterCordes的代码审查，我在使用64位模板时发现了该解决方案上的错误，但现在它可以正常运行。
这是我在运行测试时收到的输出
Creating 4 producers & 4 consumers
to flow 10.000.000 items trough the queue.
Produced: 10.743.668.245.000.000
Consumed: 5.554.289.678.184.004
Produced: 10.743.668.245.000.000
Consumed: 15.217.833.969.059.643
Produced: 10.743.668.245.000.000
Consumed: 7.380.542.769.600.801
Produced: 10.743.668.245.000.000
Consumed: 14.822.006.563.155.552
Checksum: 0 (it must be zero)