c++ - 如何避免与Helgrind产生误报？

我的线程同步“样式”似乎使helgrind掉了。这是一个重现该问题的简单程序:

#include <thread>
#include <atomic>
#include <iostream>

int main()
{
    std::atomic<bool> isReady(false);

    int i = 1;

    std::thread t([&isReady, &i]()
    {
        i = 2;
        isReady = true;
    });

    while (!isReady)
        std::this_thread::yield();

    i = 3;

    t.join();

    std::cout << i;

    return 0;
}

据我所知，上面的程序是一个格式正确的程序。但是，当我使用以下命令运行helgrind时，出现错误:

valgrind --tool=helgrind ./a.out

输出为:

==6247== Helgrind, a thread error detector
==6247== Copyright (C) 2007-2015, and GNU GPL'd, by OpenWorks LLP et al.
==6247== Using Valgrind-3.11.0 and LibVEX; rerun with -h for copyright info
==6247== Command: ./a.out
==6247==
==6247== ---Thread-Announcement------------------------------------------
==6247==
==6247== Thread #1 is the program's root thread
==6247==
==6247== ---Thread-Announcement------------------------------------------
==6247==
==6247== Thread #2 was created
==6247==    at 0x56FBB1E: clone (clone.S:74)
==6247==    by 0x4E46189: create_thread (createthread.c:102)
==6247==    by 0x4E47EC3: pthread_create@@GLIBC_2.2.5 (pthread_create.c:679)
==6247==    by 0x4C34BB7: ??? (in /usr/lib/valgrind/vgpreload_helgrind-amd64-linux.so)
==6247==    by 0x5115DC2: std::thread::_M_start_thread(std::shared_ptr<std::thread::_Impl_base>, void (*)()) (in /usr/lib/x86_64-linux-gnu/libstdc++.so.6.0.21)
==6247==    by 0x4010EF: std::thread::thread<main::{lambda()#1}>(main::{lambda()#1}&&) (in /home/arman/a.out)
==6247==    by 0x400F93: main (in /home/arman/a.out)
==6247==
==6247== ----------------------------------------------------------------
==6247==
==6247== Possible data race during read of size 1 at 0xFFF00035B by thread #1
==6247== Locks held: none
==6247==    at 0x4022C3: std::atomic<bool>::operator bool() const (in /home/arman/a.out)
==6247==    by 0x400F9F: main (in /home/arman/a.out)
==6247==
==6247== This conflicts with a previous write of size 1 by thread #2
==6247== Locks held: none
==6247==    at 0x40233D: std::__atomic_base<bool>::operator=(bool) (in /home/arman/a.out)
==6247==    by 0x40228E: std::atomic<bool>::operator=(bool) (in /home/arman/a.out)
==6247==    by 0x400F4A: main::{lambda()#1}::operator()() const (in /home/arman/a.out)
==6247==    by 0x40204D: void std::_Bind_simple<main::{lambda()#1} ()>::_M_invoke<>(std::_Index_tuple<>) (in /home/arman/a.out)
==6247==    by 0x401FA3: std::_Bind_simple<main::{lambda()#1} ()>::operator()() (in /home/arman/a.out)
==6247==    by 0x401F33: std::thread::_Impl<std::_Bind_simple<main::{lambda()#1} ()> >::_M_run() (in /home/arman/a.out)
==6247==    by 0x5115C7F: ??? (in /usr/lib/x86_64-linux-gnu/libstdc++.so.6.0.21)
==6247==    by 0x4C34DB6: ??? (in /usr/lib/valgrind/vgpreload_helgrind-amd64-linux.so)
==6247==  Address 0xfff00035b is on thread #1's stack
==6247==  in frame #1, created by main (???:)
==6247==
==6247== ----------------------------------------------------------------
==6247==
==6247== Possible data race during write of size 4 at 0xFFF00035C by thread #1
==6247== Locks held: none
==6247==    at 0x400FAE: main (in /home/arman/a.out)
==6247==
==6247== This conflicts with a previous write of size 4 by thread #2
==6247== Locks held: none
==6247==    at 0x400F35: main::{lambda()#1}::operator()() const (in /home/arman/a.out)
==6247==    by 0x40204D: void std::_Bind_simple<main::{lambda()#1} ()>::_M_invoke<>(std::_Index_tuple<>) (in /home/arman/a.out)
==6247==    by 0x401FA3: std::_Bind_simple<main::{lambda()#1} ()>::operator()() (in /home/arman/a.out)
==6247==    by 0x401F33: std::thread::_Impl<std::_Bind_simple<main::{lambda()#1} ()> >::_M_run() (in /home/arman/a.out)
==6247==    by 0x5115C7F: ??? (in /usr/lib/x86_64-linux-gnu/libstdc++.so.6.0.21)
==6247==    by 0x4C34DB6: ??? (in /usr/lib/valgrind/vgpreload_helgrind-amd64-linux.so)
==6247==    by 0x4E476F9: start_thread (pthread_create.c:333)
==6247==    by 0x56FBB5C: clone (clone.S:109)
==6247==  Address 0xfff00035c is on thread #1's stack
==6247==  in frame #0, created by main (???:)
==6247==
3==6247==
==6247== For counts of detected and suppressed errors, rerun with: -v
==6247== Use --history-level=approx or =none to gain increased speed, at
==6247== the cost of reduced accuracy of conflicting-access information
==6247== ERROR SUMMARY: 2 errors from 2 contexts (suppressed: 0 from 0)

Helgrind似乎将我的while循环作为比赛条件。我应该如何形成该程序以避免helgrind抛出误报？

最佳答案

问题在于Helgrind不了解GCC的原子内建函数，因此没有意识到它们是无种族限制的，并且在程序上强加了命令。

有很多方法可以注释您的代码以帮助Helgrind，请参见http://valgrind.org/docs/manual/hg-manual.html#hg-manual.effective-use(但是我不确定在这里如何使用它们，我已经尝试了sbabbi的显示内容，但只能解决部分问题)。

无论如何，我都会避免在繁忙的循环中屈服，这是一种糟糕的同步形式。可以使用条件变量来完成，如下所示:

#include <thread>
#include <atomic>
#include <iostream>
#include <condition_variable>

int main()
{
    bool isReady(false);
    std::mutex mx;
    std::condition_variable cv;

    int i = 1;

    std::thread t([&isReady, &i, &mx, &cv]()
    {
        i = 2;
        std::unique_lock<std::mutex> lock(mx);
        isReady = true;
        cv.notify_one();
    });

    {
        std::unique_lock<std::mutex> lock(mx);
        cv.wait(lock, [&] { return isReady; });
    }

    i = 3;

    t.join();

    std::cout << i;

    return 0;
}