【发布时间】:2020-01-13 15:50:11
【问题描述】:
考虑以下代码:
#include <mutex>
#include <thread>
#include <vector>
#include <thread>
struct counter {
counter()
{
}
size_t
get()
{
std::unique_lock<std::mutex> lock(mutex);
return 0;
}
std::mutex mutex;
};
static counter &
get_counter()
{
static counter manager;
return manager;
};
static void do_work()
{
get_counter().get();
}
int main()
{
std::vector<std::thread> v;
for (int i = 0; i < 16; i++)
v.emplace_back([]{ do_work(); });
for (auto &t : v)
t.join();
}
编译时:g++ -std=c++11 source.cpp -lpthread 并在 valgrind 下运行 valgrind --tool=drd ./a.out
我收到以下错误:
==4297== drd, a thread error detector
==4297== Copyright (C) 2006-2017, and GNU GPL'd, by Bart Van Assche.
==4297== Using Valgrind-3.15.0 and LibVEX; rerun with -h for copyright info
==4297== Command: ./a.out
==4297==
==4297== Thread 3:
==4297== Conflicting load by thread 3 at 0x0010f050 size 4
==4297== at 0x4844083: vgDrd_mutex_type (drd_pthread_intercepts.c:380)
==4297== by 0x4844083: pthread_mutex_lock_intercept (drd_pthread_intercepts.c:885)
==4297== by 0x4844083: pthread_mutex_lock (drd_pthread_intercepts.c:898)
==4297== by 0x109E72: __gthread_mutex_lock(pthread_mutex_t*) (in /home/igchor/Projects/libpmemobj-cpp/build/a.out)
==4297== by 0x109EC7: std::mutex::lock() (in /home/igchor/Projects/libpmemobj-cpp/build/a.out)
==4297== by 0x10A4A8: std::unique_lock<std::mutex>::lock() (in /home/igchor/Projects/libpmemobj-cpp/build/a.out)
==4297== by 0x10A206: std::unique_lock<std::mutex>::unique_lock(std::mutex&) (in /home/igchor/Projects/libpmemobj-cpp/build/a.out)
==4297== by 0x10A0C4: counter::get() (in /home/igchor/Projects/libpmemobj-cpp/build/a.out)
==4297== by 0x10940B: do_work() (in /home/igchor/Projects/libpmemobj-cpp/build/a.out)
==4297== by 0x109420: main::{lambda()#1}::operator()() const (in /home/igchor/Projects/libpmemobj-cpp/build/a.out)
==4297== by 0x109E06: void std::__invoke_impl<void, main::{lambda()#1}>(std::__invoke_other, main::{lambda()#1}&&) (in /home/igchor/Projects/libpmemobj-cpp/build/a.out)
==4297== by 0x109DA7: std::__invoke_result<main::{lambda()#1}>::type std::__invoke<main::{lambda()#1}>(std::__invoke_result&&, (main::{lambda()#1}&&)...) (in /home/igchor/Projects/libpmemobj-cpp/build/a.out)
==4297== by 0x109D45: void std::thread::_Invoker<std::tuple<main::{lambda()#1}> >::_M_invoke<0ul>(std::_Index_tuple<0ul>) (in /home/igchor/Projects/libpmemobj-cpp/build/a.out)
==4297== by 0x109D06: std::thread::_Invoker<std::tuple<main::{lambda()#1}> >::operator()() (in /home/igchor/Projects/libpmemobj-cpp/build/a.out)
==4297== Allocation context: BSS section of /home/igchor/Projects/libpmemobj-cpp/build/a.out
==4297== Other segment start (thread 2)
==4297== (thread finished, call stack no longer available)
==4297== Other segment end (thread 2)
==4297== (thread finished, call stack no longer available)
==4297==
==4297==
==4297== For lists of detected and suppressed errors, rerun with: -s
==4297== ERROR SUMMARY: 15 errors from 1 contexts (suppressed: 20 from 12)
这是如何初始化静态结构的问题吗?我认为这是一个误报,因为静态对象的初始化应该是线程安全的。有什么办法可以消除这个错误?
注意:
现在,当我删除 counter 结构的构造函数时,它看起来像这样:
struct counter {
size_t
get()
{
std::unique_lock<std::mutex> lock(mutex);
return 0;
}
std::mutex mutex;
};
drd 不报告任何错误。
我在 Ubuntu 19.10 上测试了所有这些:
- g++-9.2.1(和 clang++ 9.0.0)
- valgrind 3.15
【问题讨论】:
-
可能与尝试同时初始化
static counter manager;的多个线程有关,但标准要求这样做是安全的。我看不出这里显示的代码有任何明显错误。 -
哇,这听起来很不对劲。这是哪个版本的 GCC,在哪个平台上?
-
@LightnessRacesBY-SA3.0 我已经用编译器和 valgrind 的版本更新了这个问题(见问题的底部)。
-
clang++ 8.0 也会发生。默认构造函数和在产生线程之前在
main中调用get_counter都消除了错误。使用-fsanitize=thread不会给出任何警告。 Valgrind 的helgrind工具也不会报错。
标签: c++ multithreading c++11 valgrind