将执行从一个线程转移到另一个线程以实现任务并行性和逐个调用

paulotorrens 发表于 Dev

泡桐

我正在尝试在C ++中实现未来调用机制。尽管这只是一个测试代码（有些急事），但我打算在正在使用的语言的运行时中使用类似的东西来实现透明并行。

我干了我正在处理的代码，以使其变得更小，尽管它仍然很大：

#include <cstdlib>
#include <cstdio>
#include <iostream>
#include <vector>
#include <queue>
#include <future>
#include <thread>
#include <functional>
#include <type_traits>
#include <utility>
using namespace std;
using namespace std::chrono;

//------------------------------------------------------------------------------
// Simple locked printer

static std::recursive_mutex print_lock;

inline void print_() {
  return;
};

template<typename T, typename... Args>
inline void print_(T t, Args... args) {
  print_lock.lock();
  std::cout << t;
  print_(args...);
  print_lock.unlock();
};
//------------------------------------------------------------------------------

template<typename R>
class PooledTask {
  public:
    explicit PooledTask(function<R()>);

    // Possibly execute the task and return the value
    R &operator () () {

      // If we can get the lock, we're not executing
      if(lock.try_lock()) {

        // We may already have executed it
        if(done)
          goto end;

        // Otherwise, execute it now
        try {
          result = move(task());
        } catch(...) {
          // If an exception is thrown, save it for later
          eptr = current_exception();
          failed = true;
        };

        done = true;

        goto end;

      } else {

        // Wait until the task is completed
        lock.lock();

        end: {
          lock.unlock();

          // Maybe we got an exception!
          if(failed)
            rethrow_exception(eptr);

          // Otherwise, just return the result
          return result;
        };
      };
    };

  private:
    exception_ptr eptr;
    function<R()> task;
    bool done;
    bool failed;
    mutex lock;
    R result;
};

extern class TaskPool pool;

class TaskPool {
  public:
    TaskPool() noexcept: TaskPool(thread::hardware_concurrency() - 1) {
      return;
    };

    TaskPool(const TaskPool &) = delete;
    TaskPool(TaskPool &&) = delete;

    template<typename T>
    void push(PooledTask<T> *task) noexcept {

      lock_guard<mutex> guard(lock);

      builders.push([=] {
        try {
          (*task)();
        } catch(...) {
          // Ignore it here! The task will save it. :)
        };
      });

    };

    ~TaskPool() {
      // TODO: wait for all tasks to finish...
    };
  private:
    queue<thread *> threads;
    queue<function<void()>> builders;
    mutex lock;

    TaskPool(signed N) noexcept {
      while(N --> 0)
        threads.push(new thread([this, N] {
          for(;;) {

            pop_task();

          };
        }));
    };

    void pop_task() noexcept {

      lock.lock();

      if(builders.size()) {

        auto task = builders.front();

        builders.pop();

        lock.unlock();

        task();

      } else
        lock.unlock();
    };

} pool;


template<typename R>
PooledTask<R>::PooledTask(function<R()> fun):
  task(fun),
  done(false),
  failed(false)
{
  pool.push(this);
};

// Should probably return a std::shared_ptr here...
template<typename F, typename... Args>
auto byfuture(F fun, Args&&... args) noexcept ->
  PooledTask<decltype(fun(args...))> *
{

  using R = decltype(fun(args...));

  auto pooled = new PooledTask<R> {
    bind(fun, forward<Args>(args)...)
  };

  return pooled;
};


//------------------------------------------------------------------------------
#include <map>

// Get the current thread id as a simple number
static int myid() noexcept {
  static unsigned N = 0;
  static map<thread::id, unsigned> hash;
  static mutex lock;

  lock_guard<mutex> guard(lock);

  auto current = this_thread::get_id();

  if(!hash[current])
    hash[current] = ++N;

  return hash[current];
};
//------------------------------------------------------------------------------

//------------------------------------------------------------------------------
// The fibonacci test implementation
int future_fib(int x, int parent) {

  if(x < 3)
    return 1;

  print_("future_fib(", x, ")", " on thread ", myid(), \
         ", asked by thread ", parent, "\n");

  auto f1 = byfuture(future_fib, x - 1, myid());
  auto f2 = byfuture(future_fib, x - 2, myid());

  auto res = (*f1)() + (*f2)();

  delete f1;
  delete f2;

  return res;
};
//------------------------------------------------------------------------------

int main() {
  // Force main thread to get id 1
  myid();

  // Get task
  auto f = byfuture(future_fib, 8, myid());

  // Make sure it starts on the task pool
  this_thread::sleep_for(seconds(1));

  // Blocks
  (*f)();

  // Simply wait to be sure all threads are clean
  this_thread::sleep_for(seconds(2));

  //
  return EXIT_SUCCESS;
};

该程序的结果是这样的（我有一个四核，所以池中有3个线程）：

future_fib(8) on thread 2, asked by thread 1
future_fib(7) on thread 3, asked by thread 2
future_fib(6) on thread 4, asked by thread 2
future_fib(6) on thread 3, asked by thread 3
future_fib(5) on thread 4, asked by thread 4
future_fib(5) on thread 3, asked by thread 3
future_fib(4) on thread 4, asked by thread 4
future_fib(4) on thread 3, asked by thread 3
future_fib(3) on thread 4, asked by thread 4
future_fib(3) on thread 3, asked by thread 3
future_fib(3) on thread 4, asked by thread 4
future_fib(3) on thread 3, asked by thread 3
future_fib(4) on thread 4, asked by thread 4
future_fib(4) on thread 3, asked by thread 3
future_fib(3) on thread 4, asked by thread 4
future_fib(3) on thread 3, asked by thread 3
future_fib(5) on thread 3, asked by thread 3
future_fib(4) on thread 3, asked by thread 3
future_fib(3) on thread 3, asked by thread 3
future_fib(3) on thread 3, asked by thread 3