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为了试验线程清理器,我创建了一个小型 C++ 程序,该程序有意包含数据竞争。确实,tsan 确实检测到了错误,太棒了!但是我对生成的消息感到困惑......

  1. 它报告了一个写-写数据竞争,我本来预计会出现一个读写竞争。我希望find()不要写在我的容器中。如果我进一步进行小的代码调整以尝试获取 的const版本set::find(),似乎仍然存在相同的写入竞争。
  2. 它显示了 4 字节原子写入和同一地址的 8 字节写入之间的写入冲突。容器类中的同一个字段被两种不同的访问类型访问,这似乎很奇怪。

是否可以选择使用find()不写入 STL 容器的 const?

这是经过测试的 C++ 程序:

/*****************************************************************************
 * Small example with an inter-thread data race that is not obvious.
 * the error is a consequence of the non-threadsafeness of the STL containers.
 * Threading is created through portable C++11 constructs.
 * Tsan does detect the data race(?).
 *
 * Compile with one of:
 * g++-4.8 -std=c++11 -g -Wall -o race-stl11b race-stl11b.cc -pthread
 * g++-4.8 -std=c++11 -g -Wall -fsanitize=thread -fPIE -o race-stl11b-tsan race-stl11b.cc -ltsan -pie -pthread
 ******************************************************************************/

#include <iostream>
#include <thread>
#include <set>

int main()
{
    // create an empty bucket
    std::set<int> bucket;

    // Use a background task to insert value '5' in the bucket 
    std::thread t([&](){ bucket.insert(5); });

    // Check if value '3' is in the bucket (not expected :-)
    bool contains3 = bucket.find(3) != bucket.cend();
    std::cout << "Foreground find done: " << contains3 << std::endl;

    // Wait for the background thread to finish
    t.join();

    // verify that value '5' did arrive in the bucket
    bool contains5 = bucket.find(5) != bucket.cend();
    std::cout << "Background insert: " << contains5 << std::endl;

    return 0;
}

这是 tsan 输出的(部分):

WARNING: ThreadSanitizer: data race (pid=21774)                                                                                               

  Write of size 8 at 0x7d080000bfc8 by thread T1:                                                                                             
    #0 <null> <null>:0 (libtsan.so.0+0x00000001e2c0)                                                                                          
    #1 deallocate /usr/include/c++/4.8/ext/new_allocator.h:110 (exe+0x000000002a79)                                                           
    #2 deallocate /usr/include/c++/4.8/bits/alloc_traits.h:377 (exe+0x000000002962)                                                           
    #3 _M_destroy /usr/include/c++/4.8/bits/shared_ptr_base.h:417 (exe+0x00000000306b)                                                        
    #4 <null> <null>:0 (libstdc++.so.6+0x0000000b5f8a)                                                                                        

  Previous atomic write of size 4 at 0x7d080000bfc8 by main thread:
    #0 <null> <null>:0 (libtsan.so.0+0x00000000da45)
    #1 __exchange_and_add /usr/include/c++/4.8/ext/atomicity.h:49 (exe+0x000000001c9f)
    #2 __exchange_and_add_dispatch /usr/include/c++/4.8/ext/atomicity.h:82 (exe+0x000000001d56)
    #3 std::_Sp_counted_base<(__gnu_cxx::_Lock_policy)2>::_M_release() /usr/include/c++/4.8/bits/shared_ptr_base.h:141 (exe+0x00000000390d)
    #4 std::__shared_count<(__gnu_cxx::_Lock_policy)2>::~__shared_count() /usr/include/c++/4.8/bits/shared_ptr_base.h:553 (exe+0x00000000363c)
    #5 std::__shared_ptr<std::thread::_Impl_base, (__gnu_cxx::_Lock_policy)2>::~__shared_ptr() /usr/include/c++/4.8/bits/shared_ptr_base.h:810
 (exe+0x00000000351b)
    #6 std::shared_ptr<std::thread::_Impl_base>::~shared_ptr() /usr/include/c++/4.8/bits/shared_ptr.h:93 (exe+0x000000003547)
    #7 thread<main()::__lambda0> /usr/include/c++/4.8/thread:135 (exe+0x0000000020c3)
    #8 main /home/......./race-stl11b.cc:22 (exe+0x000000001e38)

感谢您的任何反馈,乔斯

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1 回答 1

1

看起来 ThreadSanitizer 在 std::thread 实现上给了你一个误报。

减少您的示例以不进行任何集合操作,如下所示:

#include <iostream>
#include <thread>
#include <set>

int main()
{
    std::set<int> bucket;
    std::thread t([&](){ /*bucket.insert(5);*/ });
    t.join();

    return 0;
}

在 ThreadSanitizer 中仍然给出相同的错误。

请注意,ThreadSanitizer 不会找到您的读写竞争条件。

于 2014-03-07T16:16:35.723 回答