c++11 - 为什么ThreadSanitizer使用此无锁示例报告比赛？

我将其简化为一个简单的独立示例。主线程排队入队1000个项目，而工作线程尝试同时出队。 ThreadSanitizer抱怨说，即使有一个获取释放内存屏障序列可以保护它们，在其中一个元素的读取和写入之间也存在竞争。

#include <atomic>
#include <thread>
#include <cassert>

struct FakeQueue
{
    int items[1000];
    std::atomic<int> m_enqueueIndex;
    int m_dequeueIndex;

    FakeQueue() : m_enqueueIndex(0), m_dequeueIndex(0) { }

    void enqueue(int x)
    {
        auto tail = m_enqueueIndex.load(std::memory_order_relaxed);
        items[tail] = x;              // <- element written
        m_enqueueIndex.store(tail + 1, std::memory_order_release);
    }

    bool try_dequeue(int& x)
    {
        auto tail = m_enqueueIndex.load(std::memory_order_acquire);
        assert(tail >= m_dequeueIndex);
        if (tail == m_dequeueIndex)
            return false;
        x = items[m_dequeueIndex];    // <- element read -- tsan says race!
        ++m_dequeueIndex;
        return true;
    }
};


FakeQueue q;

int main()
{
    std::thread th([&]() {
        int x;
        for (int i = 0; i != 1000; ++i)
            q.try_dequeue(x);
    });

    for (int i = 0; i != 1000; ++i)
        q.enqueue(i);

    th.join();
}

ThreadSanitizer输出:

==================
WARNING: ThreadSanitizer: data race (pid=17220)
  Read of size 4 at 0x0000006051c0 by thread T1:
    #0 FakeQueue::try_dequeue(int&) /home/cameron/projects/concurrentqueue/tests/tsan/issue49.cpp:26 (issue49+0x000000402bcd)
    #1 main::{lambda()#1}::operator()() const <null> (issue49+0x000000401132)
    #2 _M_invoke<> /usr/include/c++/5.3.1/functional:1531 (issue49+0x0000004025e3)
    #3 operator() /usr/include/c++/5.3.1/functional:1520 (issue49+0x0000004024ed)
    #4 _M_run /usr/include/c++/5.3.1/thread:115 (issue49+0x00000040244d)
    #5 <null> <null> (libstdc++.so.6+0x0000000b8f2f)

  Previous write of size 4 at 0x0000006051c0 by main thread:
    #0 FakeQueue::enqueue(int) /home/cameron/projects/concurrentqueue/tests/tsan/issue49.cpp:16 (issue49+0x000000402a90)
    #1 main /home/cameron/projects/concurrentqueue/tests/tsan/issue49.cpp:44 (issue49+0x000000401187)

  Location is global 'q' of size 4008 at 0x0000006051c0 (issue49+0x0000006051c0)

  Thread T1 (tid=17222, running) created by main thread at:
    #0 pthread_create <null> (libtsan.so.0+0x000000027a67)
    #1 std::thread::_M_start_thread(std::shared_ptr<std::thread::_Impl_base>, void (*)()) <null> (libstdc++.so.6+0x0000000b9072)
    #2 main /home/cameron/projects/concurrentqueue/tests/tsan/issue49.cpp:41 (issue49+0x000000401168)

SUMMARY: ThreadSanitizer: data race /home/cameron/projects/concurrentqueue/tests/tsan/issue49.cpp:26 FakeQueue::try_dequeue(int&)
==================
ThreadSanitizer: reported 1 warnings

命令行:

g++ -std=c++11 -O0 -g -fsanitize=thread issue49.cpp -o issue49 -pthread

g++版本:5.3.1

任何人都可以阐明为什么tsan认为这是一场数据竞赛吗？

更新

看来这是一个误报。为了安抚ThreadSanitizer，我添加了注释(有关支持的注释，请参见here，有关示例，请参见here)。请注意，检测是否通过宏在GCC中启用了tsan的是only recently been added，因此我现在必须手动将-D__SANITIZE_THREAD__传递给g++。

#if defined(__SANITIZE_THREAD__)
#define TSAN_ENABLED
#elif defined(__has_feature)
#if __has_feature(thread_sanitizer)
#define TSAN_ENABLED
#endif
#endif

#ifdef TSAN_ENABLED
#define TSAN_ANNOTATE_HAPPENS_BEFORE(addr) \
    AnnotateHappensBefore(__FILE__, __LINE__, (void*)(addr))
#define TSAN_ANNOTATE_HAPPENS_AFTER(addr) \
    AnnotateHappensAfter(__FILE__, __LINE__, (void*)(addr))
extern "C" void AnnotateHappensBefore(const char* f, int l, void* addr);
extern "C" void AnnotateHappensAfter(const char* f, int l, void* addr);
#else
#define TSAN_ANNOTATE_HAPPENS_BEFORE(addr)
#define TSAN_ANNOTATE_HAPPENS_AFTER(addr)
#endif

struct FakeQueue
{
    int items[1000];
    std::atomic<int> m_enqueueIndex;
    int m_dequeueIndex;

    FakeQueue() : m_enqueueIndex(0), m_dequeueIndex(0) { }

    void enqueue(int x)
    {
        auto tail = m_enqueueIndex.load(std::memory_order_relaxed);
        items[tail] = x;
        TSAN_ANNOTATE_HAPPENS_BEFORE(&items[tail]);
        m_enqueueIndex.store(tail + 1, std::memory_order_release);
    }

    bool try_dequeue(int& x)
    {
        auto tail = m_enqueueIndex.load(std::memory_order_acquire);
        assert(tail >= m_dequeueIndex);
        if (tail == m_dequeueIndex)
            return false;
        TSAN_ANNOTATE_HAPPENS_AFTER(&items[m_dequeueIndex]);
        x = items[m_dequeueIndex];
        ++m_dequeueIndex;
        return true;
    }
};

// main() is as before

现在，ThreadSanitizer在运行时很高兴。

最佳答案

ThreadSanitizer不擅长计数，它无法理解写入项目总是在读取之前进行。

ThreadSanitizer可以发现m_enqueueIndex的存储发生在加载之前，但是当items[m_dequeueIndex]时，它不了解tail > m_dequeueIndex的存储必须发生在加载之前。

关于c++11 - 为什么ThreadSanitizer使用此无锁示例报告比赛？，我们在Stack Overflow上找到一个类似的问题：https://stackoverflow.com/questions/37552866/