c++ - boost::shared_future 和 when_all 有多个延续

Question

我有一个 DAG 任务，我正在尝试使用 boost::shared_future 框架执行这些任务。

举个具体的例子，考虑图中的data flow graph。

下面是对此进行编码的尝试:

#include <iostream>

#define BOOST_THREAD_PROVIDES_FUTURE
#define BOOST_THREAD_PROVIDES_FUTURE_CONTINUATION
#define BOOST_THREAD_PROVIDES_FUTURE_WHEN_ALL_WHEN_ANY
#include <boost/thread/future.hpp>

using namespace boost;

int main() {
   shared_future<int> fa = async([]() { sleep(1); return 123; });
   shared_future<int> fb = async([]() { sleep(2); return 456; });
   shared_future<int> fc = async([]() { sleep(5); return 789; });

  auto fabc = when_all(fa,fb,fc);
  auto fx = fabc.then([](decltype(fabc)) {
    std::cout << "A,B,C has completed, computing X\n";
    return 1;
  });
  auto fax = when_all(fa,std::move(fx));
  auto fz = fax.then([](decltype(fax)) {
    std::cout << "A,X has completed, computing Z\n";
    return 2;
  });
  auto fcx = when_all(fc,std::move(fx));  // <---- BAD
  auto fy = fcx.then([](decltype(fcx)) {
    std::cout << "C,X has completed, computing Y\n";
    return 3;
  });
  fy.get();
  fz.get();
}

但是，这不起作用(显然，因为我在 fx 上调用了两次 std::move)。我想问题是 - 有没有办法让 when_all 和 then 返回“共享”类型，以便明智地执行？或者 task-DAG 的执行是否超出了 boost 的极限？

Answer 1

喜欢 T.C.也就是说，您可以通过调用 share() 成员函数来分享您的 future 。这样你就不需要移动两次:

Live On Coliru

#include <iostream>

#define BOOST_THREAD_PROVIDES_FUTURE
#define BOOST_THREAD_PROVIDES_FUTURE_CONTINUATION
#define BOOST_THREAD_PROVIDES_FUTURE_WHEN_ALL_WHEN_ANY
#include <boost/thread/future.hpp>

using namespace boost;
using boost::this_thread::sleep_for;
using boost::chrono::milliseconds;

int main() {
    shared_future<int> fa = async([]() { sleep_for(milliseconds(100)); return 123; });
    shared_future<int> fb = async([]() { sleep_for(milliseconds(200)); return 456; });
    shared_future<int> fc = async([]() { sleep_for(milliseconds(500)); return 789; });

    auto fabc = when_all(fa, fb, fc);

    auto fx   = fabc
        .then([](decltype(fabc)) { std::cout << "A,B,C has completed, computing X\n"; return 1; })
        .share();
    auto fax  = when_all(fa, fx);

    auto fz   = fax
        .then([](decltype(fax)) { std::cout << "A,X has completed, computing Z\n"; return 2; })
        .share();
    auto fcx  = when_all(fc, fx);

    auto fy   = fcx
        .then([](decltype(fcx)) { std::cout << "C,X has completed, computing Y\n"; return 3; })
        .share();

    fy.get();
    fz.get();
}

打印

A,B,C has completed, computing X
C,X has completed, computing Y
A,X has completed, computing Z