condition_variable.notify 是同步点吗? [英] Is condition_variable.notify a synchronization point?

问题描述

假设我有这样的事情:

bool signalled = false;
std::condition_variable cv;
void thread1() {
  while (true) {
    std::unique_lock l(mutex);
    cv.wait_until(l, [] { return signalled; });
    return;
  }
}

void thread2...N() {
  signalled = true;
  cv.notify_all();
}

这被认为是线程安全的吗?布尔值可能在许多线程中设置为 true 以中断线程 1.

Is that considered thread-safe? The boolean may be set to true in many threads to interrupt thread1.

如果不是线程安全的，我正在寻找竞争条件的描述，以便我可以更好地理解潜在问题并填补知识空白.

If not thread-safe I’m looking for a description of what the race condition is so I can better understand the underlying issue and fill in a knowledge gap.

推荐答案

在没有同步的情况下写入非原子变量是 UB.但是，将 signaled 设为 atomic 并不能解决问题.

Writing to a non-atomic variable without synchronization is UB. However, making signalled an atomic<bool> will not solve the problem.

C++ 参考关于std::condition_variable 读:

即使共享变量是原子的，也必须在互斥锁下进行修改，才能将修改正确发布到等待线程.

Even if the shared variable is atomic, it must be modified under the mutex in order to correctly publish the modification to the waiting thread.

你应该这样做:

bool signalled = false;

void thread2...N()
{
    std::unique_lock l(mutex);
    signalled = true;
    cv.notify_all();
}

相关问题:

• 互斥保护 std::condition_variable
• 共享如果没有在互斥锁下修改原子变量，则不会正确发布
• 为什么我需要获取锁来修改共享的原子"?通知condition_variable前的变量

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