我尝试从文档页面https://en.cppreference.com/w/cpp/thread/call_once运行该示例,但未按预期运行。它被无限卡住。我想知道为什么会发生这种情况,或者这仅仅是与某些特定编译器版本相关的错误。这就是我用来运行程序https://repl.it/repls/UtterJubilantArchitects
的方式#include <iostream>
#include <thread>
#include <mutex>
std::once_flag flag1, flag2;
void simple_do_once()
{
std::call_once(flag1, [](){ std::cout << "Simple example: called once\n"; });
}
void may_throw_function(bool do_throw)
{
if (do_throw) {
std::cout << "throw: call_once will retry\n"; // this may appear more than once
throw std::exception();
}
std::cout << "Didn't throw, call_once will not attempt again\n"; // guaranteed once
}
void do_once(bool do_throw)
{
try {
std::call_once(flag2, may_throw_function, do_throw);
}
catch (...) {
}
}
int main()
{
std::thread st1(simple_do_once);
std::thread st2(simple_do_once);
std::thread st3(simple_do_once);
std::thread st4(simple_do_once);
st1.join();
st2.join();
st3.join();
st4.join();
std::thread t1(do_once, true);
std::thread t2(do_once, true);
std::thread t3(do_once, false);
std::thread t4(do_once, true);
t1.join();
t2.join();
t3.join();
t4.join();
}
答案 0 :(得分:6)
此行为是实现错误。 call_onse(通过pthread_once)应该使用 int pthread_mutex_lock(pthread_mutex_t )*和 int pthread_mutex_unlock(pthread_mutex_t )*,以及它们之间的代码,它们都不安全。
链接到一些相关的错误:https://gcc.gnu.org/bugzilla/show_bug.cgi?id=66146, sourceware.org/bugzilla/show_bug.cgi?id=18435, austingroupbugs.net/view.php?id=863#c2619
实施细节:
这是来自gnu.org的pthread实现
int
__pthread_once (pthread_once_t *once_control, void (*init_routine) (void))
{
__memory_barrier ();
if (once_control->__run == 0)
{
__pthread_spin_lock (&once_control->__lock);
if (once_control->__run == 0)
{
init_routine ();
__memory_barrier ();
once_control->__run = 1;
}
__pthread_spin_unlock (&once_control->__lock);
}
return 0;
}
这是来自g ++ 7.4.0程序包(ubuntu 18.4.0)的call_once实现: (我们有有效的_GLIBCXX_HAVE_TLS分支)
/// call_once
template<typename _Callable, typename... _Args>
void
call_once(once_flag& __once, _Callable&& __f, _Args&&... __args)
{
// _GLIBCXX_RESOLVE_LIB_DEFECTS
// 2442. call_once() shouldn't DECAY_COPY()
auto __callable = [&] {
std::__invoke(std::forward<_Callable>(__f),
std::forward<_Args>(__args)...);
};
#ifdef _GLIBCXX_HAVE_TLS
__once_callable = std::__addressof(__callable);
__once_call = []{ (*(decltype(__callable)*)__once_callable)(); };
#else
unique_lock<mutex> __functor_lock(__get_once_mutex());
__once_functor = __callable;
__set_once_functor_lock_ptr(&__functor_lock);
#endif
int __e = __gthread_once(&__once._M_once, &__once_proxy);
#ifndef _GLIBCXX_HAVE_TLS
if (__functor_lock)
__set_once_functor_lock_ptr(0);
#endif
#ifdef __clang_analyzer__
// PR libstdc++/82481
__once_callable = nullptr;
__once_call = nullptr;
#endif
if (__e)
__throw_system_error(__e);
}
__ once_proxy是:
extern "C"
{
void __once_proxy()
{
#ifndef _GLIBCXX_HAVE_TLS
function<void()> __once_call = std::move(__once_functor);
if (unique_lock<mutex>* __lock = __get_once_functor_lock_ptr())
{
// caller is using new ABI and provided lock ptr
__get_once_functor_lock_ptr() = 0;
__lock->unlock();
}
else
__get_once_functor_lock().unlock(); // global lock
#endif
__once_call();
}
}
_GLIBCXX_END_NAMESPACE_VERSION
} // namespace std
这是pthread_once的实现:
struct __pthread_once
{
int __run;
__pthread_spinlock_t __lock;
};
和__pthread_spinlock_t:
typedef __volatile int __pthread_spinlock_t;
__ pthread_spin_lock是
的typedefvoid
__spin_lock_solid (spin_lock_t *lock)
{
while (__spin_lock_locked (lock) || ! __spin_try_lock (lock))
/* Yield to another thread (system call). */
__swtch_pri (0);
}
就个人而言,当互斥锁解锁保护不受例外影响时,我在这些区域看不到。我们在内部仔细检查了锁和Callable对象的调用。
我认为,除了Callable或带有lock_guard / unique_lock的复查锁(如果您确定读取变量的原子性)之外,不能将其用作解决方案。我一直关注walnut的解决方案, 安装了libc ++-dev,libc ++ abi-dev并使用
编译了此代码clang++ -stdlib=libc++ -lc++abi
效果很好。