c++ - std::thread::join是否保证写入可见性

Question

据说Std::thread::join与所连接的线程“同步”，但是同步并不能说明副作用的可见性，它仅控制可见性的顺序，即。在以下示例中:

int g_i = 0;
int main()
{
  auto fn = [&] {g_i = 1;};
  std::thread t1(fn);
  t1.join();
  return g_i;
}

在c++标准中，我们是否对保证此程序将始终返回1？

Answer 1

[thread.thread.member]:

void join();
Effects: Blocks until the thread represented by *this has completed.
Synchronization: The completion of the thread represented by *this synchronizes with the corresponding successful join() return.

由于线程执行的完成与thread::join的返回值同步，因此线程inter-thread happens before的返回完成了:

An evaluation A inter-thread happens before an evaluation B if
— A synchronizes with B

因此happens before它:

An evaluation A happens before an evaluation B (or, equivalently, B happens after A) if:
— A inter-thread happens before B

由于(线程间)发生在传递性之前(让我跳过复制以粘贴整个线程间定义来显示此行为)，因此在线程完成之前发生的所有事情，包括将值1写入g_i，都会发生从thread::join返回之前。继而从thread::join返回的结果发生在读取g_i中的return g_i;的值之前，这仅仅是因为thread::join的调用在return g_i;之前进行了排序。再次，使用可传递性，我们确定在非主线程中将1写入g_i的操作发生在主线程中g_i中的return g_i;的读取之前。
对于1中的g_i的读取，将g_i写入return g_i;是visible side effect:

A visible side effect A on a scalar object or bit-field M with respect to a value computation B of M satisfies the conditions:
— A happens before B and
— there is no other side effect X to M such that A happens before X and X happens before B.
The value of a non-atomic scalar object or bit-field M, as determined by evaluation B, shall be the value stored by the visible side effect A.

最后一句话的重点是我的，它保证了从g_i中的return g_i;读取的值将是1。