Question

我有以下代码，它会异步触发在任务向量上推送的每个已完成任务的回调

#define BOOST_THREAD_PROVIDES_FUTURE
#define BOOST_THREAD_PROVIDES_FUTURE_CONTINUATION

#include <fstream>
#include <iostream>
#include <random>
#include <thread>
#include <vector>

#include <boost/thread/future.hpp>

using vbf = std::vector<boost::future<std::pair<std::string, int>>>;

void processFile(vbf& v, const std::string& f, int offset)
{
  v.emplace_back(boost::async([offset, &f]() {
    std::cout << "Starting " << f << '\n';
    std::ofstream ofs(f);
    const int max = offset * 1000000;
    for (int i = 0; i < max; ++i) ofs << i;
    // line below wouldn't present the real problem
    //std::this_thread::sleep_for(std::chrono::milliseconds(max));
    return std::make_pair(f, max);
  }));
}

void printProgress(vbf& tasks)
{
  auto it = boost::wait_for_any(tasks.begin(), tasks.end());
  while (it != tasks.end())
  {
    const auto res = it->get();
    std::cout << "processed file \"" << res.first << "\" with " << res.second << " records\n";
    tasks.erase(it);
    it = boost::wait_for_any(tasks.begin(), tasks.end());
  }
}

int main()
{
  std::vector<std::string> vs{
      "file1",  "file2",  "file3",  "file4",  "file5",  "file6",  "file7",  "file8",  "file9",
      "file10", "file11", "file12", "file13", "file14", "file15", "file16", "file17", "file18",
      "file19", "file20", "file21", "file22", "file23", "file24", "file25", "file26", "file27",
      "file28", "file29", "file30", "file31", "file32", "file33", "file34", "file35", "file36",
      "file37", "file38", "file39", "file40", "file41", "file42", "file43", "file44", "file45",
      "file46", "file47", "file48", "file49", "file50"};

  vbf v;
  v.reserve(vs.size());

  std::random_device rd;
  std::mt19937 gen(rd());
  std::uniform_int_distribution<> dis(1, 10);

  for (const auto& f : vs) processFile(v, f, dis(gen));

  printProgress(v);
}

通过写入来自std::vector<std::string> vs的文件来“模拟”加载（有sleep()的评论，但这不会出现真正的问题，因为CPU上没有实际负载）

我的问题是如何/通过什么方法可以正确地安排/排队在这里完成的工作，这样我就不会得到争用，线程在CPU上“打架”等等？

我意识到这是一个非常重要的问题，它可能只有少数解决方案/方法。

在线IDE http://melpon.org/wandbox/permlink/aME5cUecefZaBof4

Answer 1

为了避免争用，你要做的第一件事就是不要开始/运行更多线程（同时），而不是核心（超线程）来执行它们。

考虑到线程数，如何在boost :: wait_for_any中正确安排任务？

1 个答案: