Question

我正在通过 Anthony Williams 阅读C++ Concurrency in Action。在关于设计并发代码的章节中，有std::for_each algorihtm的并行版本。以下是本书中稍微修改过的代码：

join_thread.hpp

#pragma once

#include <vector>
#include <thread>

class join_threads
{
public:
  explicit join_threads(std::vector<std::thread>& threads)
    : threads_(threads) {}

  ~join_threads()
  {
    for (size_t i = 0; i < threads_.size(); ++i)
    {
      if(threads_[i].joinable())
      {
        threads_[i].join();
      }
    }
  }

private:
  std::vector<std::thread>& threads_;
};

parallel_for_each.hpp

#pragma once

#include <future>
#include <algorithm>

#include "join_threads.hpp"

template<typename Iterator, typename Func>
void parallel_for_each(Iterator first, Iterator last, Func func)
{
  const auto length = std::distance(first, last);
  if (0 == length) return;

  const auto min_per_thread = 25u;
  const unsigned max_threads = (length + min_per_thread - 1) / min_per_thread;

  const auto hardware_threads = std::thread::hardware_concurrency();

  const auto num_threads= std::min(hardware_threads != 0 ?
        hardware_threads : 2u, max_threads);

  const auto block_size = length / num_threads;

  std::vector<std::future<void>> futures(num_threads - 1);
  std::vector<std::thread> threads(num_threads-1);
  join_threads joiner(threads);

  auto block_start = first;
  for (unsigned i = 0; i < num_threads - 1; ++i)
  {
    auto block_end = block_start;
    std::advance(block_end, block_size);
    std::packaged_task<void (void)> task([block_start, block_end, func]()
    {
      std::for_each(block_start, block_end, func);
    });
    futures[i] = task.get_future();
    threads[i] = std::thread(std::move(task));
    block_start = block_end;
  }

  std::for_each(block_start, last, func);

  for (size_t i = 0; i < num_threads - 1; ++i)
  {
    futures[i].get();
  }
}

我使用以下程序将其与std::for_each的顺序版本进行基准测试：

的main.cpp

#include <iostream>
#include <random>
#include <chrono>

#include "parallel_for_each.hpp"

using namespace std;

constexpr size_t ARRAY_SIZE = 500'000'000;
typedef std::vector<uint64_t> Array;

template <class FE, class F>
void test_for_each(const Array& a, FE fe, F f, atomic<uint64_t>& result)
{
  auto time_begin = chrono::high_resolution_clock::now();
  result = 0;
  fe(a.begin(), a.end(), f);
  auto time_end = chrono::high_resolution_clock::now();

  cout << "Result = " << result << endl;
  cout << "Time: " << chrono::duration_cast<chrono::milliseconds>(
            time_end - time_begin).count() << endl;
}

int main()
{
  random_device device;
  default_random_engine engine(device());
  uniform_int_distribution<uint8_t> distribution(0, 255);

  Array a;
  a.reserve(ARRAY_SIZE);

  cout << "Generating array ... " << endl;
  for (size_t i = 0; i < ARRAY_SIZE; ++i)
    a.push_back(distribution(engine));

  atomic<uint64_t> result;
  auto acc = [&result](uint64_t value) { result += value; };

  cout << "parallel_for_each ..." << endl;
  test_for_each(a, parallel_for_each<Array::const_iterator, decltype(acc)>, acc, result);
  cout << "for_each ..." << endl;
  test_for_each(a, for_each<Array::const_iterator, decltype(acc)>, acc, result);

  return 0;
}

我机器上算法的并行版本比顺序版本慢两倍：

parallel_for_each ...
Result = 63750301073
Time: 5448
for_each ...
Result = 63750301073
Time: 2496

我在 Intel（R）Core（TM）i3-6100 CPU @ 3.70GHz <上运行的 Ubuntu Linux 上使用 GCC 6.2 编译器< /强>

如何解释这种行为？这是因为在线程和缓存ping-pong之间共享atomic<uint64_t>变量？

我使用性能分别对其进行了分析。对于并行版本，统计数据如下：

1137982167 cache-references 247652893 cache-misses # 21,762 % of all cache refs 60868183996 cycles 27409239189 instructions # 0,45 insns per cycle 3287117194 branches 80895 faults 4 migrations

顺序之一：

402791485 cache-references 246561299 cache-misses # 61,213 % of all cache refs 40284812779 cycles 26515783790 instructions # 0,66 insns per cycle 3188784664 branches 48179 faults 3 migrations

很明显，并行版本会产生更多的缓存引用，循环和错误，但为什么呢？

Answer 1

你共享相同的result变量：所有线程都在atomic<uint64_t> result上累积，颠倒了缓存！

每次线程写入result时，其他核心中的所有缓存都会失效：这会导致缓存行争用。

更多信息：

"Sharing Is the Root of All Contention"

[...]要写入内存位置，核心必须另外拥有包含该位置的缓存行的独占所有权。虽然一个核心是独占使用的，但是所有其他尝试写入相同内存位置的核心必须等待并轮流 - 也就是说，它们必须串行运行。从概念上讲，它就好像每个缓存行都受到硬件互斥锁的保护，其中只有一个核心可以同时保存该缓存行上的硬件锁。
This article on "false sharing"涵盖了类似的问题，更深入地解释了缓存中发生的事情。

我对您的程序进行了一些修改并获得了以下结果（在i7-4770K [8个主题+超线程]的机器上）：

Generating array ...
parallel_for_each ...
Result = 63748111806
Time: 195
for_each ...
Result = 63748111806
Time: 2727

并行版本比串行版本大约快92％。

std::future和std::packaged_task是重量级抽象。在这种情况下，std::experimental::latch就足够了。
每个任务都发送到线程池这个最小化线程创建开销。
每个任务都有自己的累加器。 消除了分享。

代码可用here on my GitHub。它使用了一些个人依赖项，但无论如何都应该理解这些变化。

以下是最重要的变化：

// A latch is being used instead of a vector of futures.
ecst::latch l(num_threads - 1);

l.execute_and_wait_until_zero([&]
{
    auto block_start = first;
    for (unsigned i = 0; i < num_threads - 1; ++i)
    {
        auto block_end = block_start;
        std::advance(block_end, block_size);

        // `p` is a thread pool.
        // Every task posted in the thread pool has its own `tempacc` accumulator.
        p.post([&, block_start, block_end, tempacc = 0ull]() mutable
        {
            // The task accumulator is filled up...
            std::for_each(block_start, block_end, [&tempacc](auto x){ tempacc += x; });

            // ...and then the atomic variable is incremented ONCE.
            func(tempacc);
            l.decrement_and_notify_all();
        });

        block_start = block_end;
    }

    // Same idea here: accumulate to local non-atomic counter, then
    // add the partial result to the atomic counter ONCE.
    auto tempacc2 = 0ull;
    std::for_each(block_start, last, [&tempacc2](auto x){ tempacc2 += x; });
    func(tempacc2);
});

并行for_each比std :: for_each慢两倍多

1 个答案: