我不明白,为什么没有release sequence
会出问题,如果我们在下面的示例中有 2 个线程。我们对原子变量 count
只有 2 个操作. count
如输出所示,按顺序递减。
来自安东尼威廉姆斯的 C++ Concurrency in Action:
I mentioned that you could get a
synchronizes-with relationship
between astore
to an atomic variable and aload
of that atomic variable from another thread, even when there’s a sequence ofread-modify-write
operations between thestore
and theload
, provided all the operations are suitably tagged. If the store is tagged withmemory_order_release
,memory_order_acq_rel
, ormemory_order_seq_cst
, and the load is tagged withmemory_order_consume
,memory_order_acquire
, ormemory_order_seq_cst
, and each operation in the chain loads the value written by the previous operation, then the chain of operations constitutes a release sequence and the initial storesynchronizes-with
(formemory_order_acquire
ormemory_order_seq_cst
) or isdependency-ordered-before
(formemory_order_consume
) the final load. Any atomic read-modify-write operations in the chain can have any memory ordering (evenmemory_order_relaxed
).To see what this means (release sequence) and why it’s important, consider an
atomic<int>
being used as a count of the number of items in a shared queue, as in the following listing.One way to handle things would be to have the thread that’s producingthe data store the items in a shared buffer and then do
count.store(number_of_items, memory_order_release)
#1 to let the other threads know that data is available. The threads consuming the queue items might then docount.fetch_sub(1,memory_ order_acquire)
#2 to claim an item from the queue, prior to actually reading the shared buffer #4. Once the count becomes zero, there are no more items, and the thread must wait #3.
#include <atomic>
#include <thread>
#include <vector>
#include <iostream>
#include <mutex>
std::vector<int> queue_data;
std::atomic<int> count;
std::mutex m;
void process(int i)
{
std::lock_guard<std::mutex> lock(m);
std::cout << "id " << std::this_thread::get_id() << ": " << i << std::endl;
}
void populate_queue()
{
unsigned const number_of_items = 20;
queue_data.clear();
for (unsigned i = 0;i<number_of_items;++i)
{
queue_data.push_back(i);
}
count.store(number_of_items, std::memory_order_release); //#1 The initial store
}
void consume_queue_items()
{
while (true)
{
int item_index;
if ((item_index = count.fetch_sub(1, std::memory_order_acquire)) <= 0) //#2 An RMW operation
{
std::this_thread::sleep_for(std::chrono::milliseconds(500)); //#3
continue;
}
process(queue_data[item_index - 1]); //#4 Reading queue_data is safe
}
}
int main()
{
std::thread a(populate_queue);
std::thread b(consume_queue_items);
std::thread c(consume_queue_items);
a.join();
b.join();
c.join();
}
输出(VS2015):
id 6836: 19
id 6836: 18
id 6836: 17
id 6836: 16
id 6836: 14
id 6836: 13
id 6836: 12
id 6836: 11
id 6836: 10
id 6836: 9
id 6836: 8
id 13740: 15
id 13740: 6
id 13740: 5
id 13740: 4
id 13740: 3
id 13740: 2
id 13740: 1
id 13740: 0
id 6836: 7
If there’s one consumer thread, this is fine; the
fetch_sub()
is a read, withmemory_order_acquire
semantics, and the store hadmemory_order_release
semantics, so the store synchronizes-with the load and the thread can read the item from the buffer.If there are two threads reading, the second
fetch_sub()
will see the value written by the first and not the value written by the store. Without the rule about therelease sequence
, this second thread wouldn’t have ahappens-before relationship
with the first thread, and it wouldn’t be safe to read the shared buffer unless the firstfetch_sub()
also hadmemory_order_release
semantics, which would introduce unnecessary synchronization between the two consumer threads. Without therelease sequence
rule ormemory_order_release
on thefetch_sub
operations, there would be nothing to require that the stores to thequeue_data
were visible to the second consumer, and you would have a data race.
他什么意思?两个线程都应该看到
count
的值是 20
?但在我的输出 count
在线程中依次递减。Thankfully, the first
fetch_sub()
does participate in the release sequence, and so thestore()
synchronizes-with the secondfetch_sub()
. There’s still no synchronizes-with relationship between the two consumer threads. This is shown in figure 5.7. The dotted lines in figure 5.7 show the release sequence, and the solid lines show thehappens-before relationships
最佳答案
What does he mean? That both threads should see the value of count is 20? But in my output count is sequently decremented in threads.
不,他没有。全部修改为
count
是原子的,所以在给定的代码中,两个读取器线程总是会看到不同的值。他在谈论释放顺序规则的含义,即当给定线程执行
release
时存储,然后执行的其他多个线程 acquire
相同位置的负载形成一个释放序列,其中每个后续acquire
加载与存储线程有先发生的关系(即存储的完成发生在加载之前)。这意味着读取器线程中的加载操作是与写入器线程的同步点,写入器中存储之前的所有内存操作都必须完成并在其相应加载完成时在读取器中可见。他是说如果没有这个规则,那么只有第一个线程会同步到作者。因此,第二个线程在访问
queue
时会发生数据竞争。 (注意:不是 count
,它无论如何都受到原子访问的保护)。理论上,对数据的内存操作发生在 store
之前。在 count
只有在 count
上执行自己的加载操作后,读取器线程编号 2 才能看到它.发布顺序规则确保这不会发生。总之:发布顺序规则确保多个线程可以在单个存储上同步它们的负载。有问题的同步是对数据的内存访问,而不是被同步的实际原子变量(由于是原子的,因此无论如何都保证同步)。
请注意在此处添加:在大多数情况下,这些类型的问题仅与对重新排序其内存操作放松的 CPU 架构有关。 Intel 架构不是其中之一:它是强排序的,并且只有少数非常特殊的情况可以对内存操作进行重新排序。这些细微差别大多只在谈论其他架构时才有意义,例如 ARM 和 PowerPC。
关于c++ - "release sequence"是什么意思?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/38565650/