我有一个非常简单的队列实现，包装一个固定的数组。 它包含peek，enqueue和dequeue。 如果peek返回一个引用，我发现它最终将返回冲突的结果（冲突的结果意味着它将返回2个不同的值而没有任何插入的出列或排队）。 显然，如果该引用被保留并修改，可能会发生这种情况，但据我所知，事实并非如此。 实际上，再次调用peek会给出预期的结果。
下面是Windows线程和互斥锁的代码。 我也在Linux上使用pthreads尝试了相同的结果。 我显然不明白的东西......我已经转储了可执行文件并找到了返回引用的唯一区别，或者是在取消引用内存位置时的唯一区别。 例如：
如果返回引用，则peek包含： lea eax,[edx+ecx*4+8] 然后在消费者线程中： cmp dword ptr [eax],1
但是，如果返回一个值，则peek包含： mov eax,dword ptr [edx+ecx*4+8] 然后在消费者线程中： cmp eax,1
谢谢！
#include <iostream> #include <windows.h> typedef void *(thread_func_type)(void *); void start_thread(HANDLE &thread, thread_func_type *thread_func, void *arg) { DWORD id; thread = CreateThread(NULL, 0, (LPTHREAD_START_ROUTINE)thread_func, arg, 0, &id); if (thread == NULL) { std::cerr << "ERROR: failed to create thread\n"; ::exit(1); } } void join_thread(HANDLE &thread) { WaitForSingleObject(thread, INFINITE); } class ScopedMutex { HANDLE &mutex; public: ScopedMutex(HANDLE &mutex_) : mutex(mutex_) { WORD result = WaitForSingleObject(mutex, INFINITE); if (result != WAIT_OBJECT_0) { std::cerr << "ERROR: failed to lock mutex\n"; ::exit(1); } }; ~ScopedMutex() { ReleaseMutex(mutex); }; }; template <typename T, unsigned depth> class Queue { unsigned head, tail; bool full; T data[depth]; HANDLE mutex; public: Queue() : head(0), tail(0), full(false) { mutex = CreateMutex(NULL, 0, NULL); if (mutex == NULL) { std::cerr << "ERROR: could not create mutex.\n"; ::exit(1); } }; T &peek() { while (true) { { ScopedMutex local_lock(mutex); if (full || (head != tail)) return data[tail]; } Sleep(0); } }; void enqueue(const T &t) { while (true) { { ScopedMutex local_lock(mutex); if (!full) { data[head++] = t; head %= depth; full = (head == tail); return; } } Sleep(0); } }; void dequeue() { while (true) { { ScopedMutex local_lock(mutex); if (full || (head != tail)) { ++tail; tail %= depth; full = false; return; } } Sleep(0); } }; }; template <unsigned num_vals, int val, unsigned depth> void * producer(void *arg) { Queue<int, depth> &queue = *static_cast<Queue<int, depth> *>(arg); for (unsigned i = 0; i < num_vals; ++i) { queue.enqueue(val); } std::cerr << "producer " << val << " exiting.\n"; return NULL; } template <unsigned num_vals, int val, unsigned depth> void * consumer(void *arg) { Queue<int, depth> &queue = *static_cast<Queue<int, depth> *>(arg); for (unsigned i = 0; i < num_vals; ++i) { while (queue.peek() != val) Sleep(0); if (queue.peek() != val) { std::cerr << "ERROR: (" << val << ", " << queue.peek() << ")" << std::endl; std::cerr << "But peeking again gives the right value " << queue.peek() << std::endl; ::exit(1); } queue.dequeue(); } return NULL; } int main(int argc, char *argv[]) { const unsigned depth = 10; const unsigned num_vals = 100000; Queue<int, depth> queue; HANDLE p1, p2, c1, c2; start_thread(p1, producer<num_vals, 1, depth>, &queue); start_thread(p2, producer<num_vals, 2, depth>, &queue); start_thread(c1, consumer<num_vals, 1, depth>, &queue); start_thread(c2, consumer<num_vals, 2, depth>, &queue); join_thread(p1); join_thread(p2); join_thread(c1); join_thread(c2); }I have a very simple queue implementation that wraps a fixed array. It contains peek, enqueue, and dequeue. If peek returns a reference, I've found that it will return conflicting results eventually (conflicting results meaning it will return 2 different values without any intervening dequeues or enqueues). Obviously, this could happen if that reference is held onto and modified, but as far as I can tell, it isn't. In fact, calling peek again gives the expected result.
Below is the code with Windows threading and mutexes. I've also tried this using pthreads on Linux with the same result. I obviously don't understand something... I've dumped the executable and find the only difference between returning a reference or a value is when the memory location is dereferenced. For example:
If a reference is returned, peek contains: lea eax,[edx+ecx*4+8] And then in the consumer thread: cmp dword ptr [eax],1
But, if a value is returned, peek contains: mov eax,dword ptr [edx+ecx*4+8] And then in the consumer thread: cmp eax,1
Thanks!
#include <iostream> #include <windows.h> typedef void *(thread_func_type)(void *); void start_thread(HANDLE &thread, thread_func_type *thread_func, void *arg) { DWORD id; thread = CreateThread(NULL, 0, (LPTHREAD_START_ROUTINE)thread_func, arg, 0, &id); if (thread == NULL) { std::cerr << "ERROR: failed to create thread\n"; ::exit(1); } } void join_thread(HANDLE &thread) { WaitForSingleObject(thread, INFINITE); } class ScopedMutex { HANDLE &mutex; public: ScopedMutex(HANDLE &mutex_) : mutex(mutex_) { WORD result = WaitForSingleObject(mutex, INFINITE); if (result != WAIT_OBJECT_0) { std::cerr << "ERROR: failed to lock mutex\n"; ::exit(1); } }; ~ScopedMutex() { ReleaseMutex(mutex); }; }; template <typename T, unsigned depth> class Queue { unsigned head, tail; bool full; T data[depth]; HANDLE mutex; public: Queue() : head(0), tail(0), full(false) { mutex = CreateMutex(NULL, 0, NULL); if (mutex == NULL) { std::cerr << "ERROR: could not create mutex.\n"; ::exit(1); } }; T &peek() { while (true) { { ScopedMutex local_lock(mutex); if (full || (head != tail)) return data[tail]; } Sleep(0); } }; void enqueue(const T &t) { while (true) { { ScopedMutex local_lock(mutex); if (!full) { data[head++] = t; head %= depth; full = (head == tail); return; } } Sleep(0); } }; void dequeue() { while (true) { { ScopedMutex local_lock(mutex); if (full || (head != tail)) { ++tail; tail %= depth; full = false; return; } } Sleep(0); } }; }; template <unsigned num_vals, int val, unsigned depth> void * producer(void *arg) { Queue<int, depth> &queue = *static_cast<Queue<int, depth> *>(arg); for (unsigned i = 0; i < num_vals; ++i) { queue.enqueue(val); } std::cerr << "producer " << val << " exiting.\n"; return NULL; } template <unsigned num_vals, int val, unsigned depth> void * consumer(void *arg) { Queue<int, depth> &queue = *static_cast<Queue<int, depth> *>(arg); for (unsigned i = 0; i < num_vals; ++i) { while (queue.peek() != val) Sleep(0); if (queue.peek() != val) { std::cerr << "ERROR: (" << val << ", " << queue.peek() << ")" << std::endl; std::cerr << "But peeking again gives the right value " << queue.peek() << std::endl; ::exit(1); } queue.dequeue(); } return NULL; } int main(int argc, char *argv[]) { const unsigned depth = 10; const unsigned num_vals = 100000; Queue<int, depth> queue; HANDLE p1, p2, c1, c2; start_thread(p1, producer<num_vals, 1, depth>, &queue); start_thread(p2, producer<num_vals, 2, depth>, &queue); start_thread(c1, consumer<num_vals, 1, depth>, &queue); start_thread(c2, consumer<num_vals, 2, depth>, &queue); join_thread(p1); join_thread(p2); join_thread(c1); join_thread(c2); }最满意答案
Peek将引用返回到数组的中间，而其他线程正在主动修改该内存。 从该引用中读取任何属性将是未定义的行为。 你不能偷看里面，你的出队应该删除元素并返回副本。
Peek returns a reference into the middle of array, while other threads are actively modifying that memory. Reading any property from that reference will be undefined behavior. You can't peek inside, your dequeue should remove the element and return a copy.
更多推荐
发布评论