How can I store generic packaged_tasks in a container?

Question

I'm trying to take a 'task' in the style of std::async and store it in a container. I'm having to jump through hoops to achieve it, but I think there must be a better way.

std::vector<std::function<void()>> mTasks;

template<class F, class... Args>
std::future<typename std::result_of<typename std::decay<F>::type(typename std::decay<Args>::type...)>::type>
push(F&& f, Args&&... args)
{
    auto func = std::make_shared<std::packaged_task<typename std::result_of<typename std::decay<F>::type(typename std::decay<Args>::type...)>::type()>>(std::bind(std::forward<F>(f), std::forward<Args>(args)...));
    auto future = func->get_future();

    // for some reason I get a compilation error in clang if I get rid of the `=, ` in this capture:
    mTasks.push_back([=, func = std::move(func)]{ (*func)(); });

    return future;
}

So I'm using bind -> packaged_task -> shared_ptr -> lambda -> function. How can I do this better/more optimally? It would certainly be easier if there was a std::function which could take a non-copyable but moveable task. Can I std::forward args into the capture of a lambda, or do I have to use bind?

Answer 1

There is no kill like overkill.

Step 1: write a SFINAE friendly std::result_of and a function to help calling via tuple:

namespace details {
  template<size_t...Is, class F, class... Args>
  auto invoke_tuple( std::index_sequence<Is...>, F&& f, std::tuple<Args>&& args)
  {
    return std::forward<F>(f)( std::get<Is>(std::move(args)) );
  }
  // SFINAE friendly result_of:
  template<class Invocation, class=void>
  struct invoke_result {};
  template<class T, class...Args>
  struct invoke_result<T(Args...), decltype( void(std::declval<T>()(std::declval<Args>()...)) ) > {
    using type = decltype( std::declval<T>()(std::declval<Args>()...) );
  };
  template<class Invocation, class=void>
  struct can_invoke:std::false_type{};
  template<class Invocation>
  struct can_invoke<Invocation, decltype(void(std::declval<
    typename invoke_result<Inocation>::type
  >()))>:std::true_type{};
}

template<class F, class... Args>
auto invoke_tuple( F&& f, std::tuple<Args>&& args)
{
  return details::invoke_tuple( std::index_sequence_for<Args...>{}, std::forward<F>(f), std::move(args) );
}

// SFINAE friendly result_of:
template<class Invocation>
struct invoke_result:details::invoke_result<Invocation>{};
template<class Invocation>
using invoke_result_t = typename invoke_result<Invocation>::type;
template<class Invocation>
struct can_invoke:details::can_invoke<Invocation>{};

We now have invoke_result_t<A(B,C)> which is a SFINAE friendly result_of_t<A(B,C)> and can_invoke<A(B,C)> which just does the check.

Next, write a move_only_function, a move-only version of std::function:

namespace details {
  template<class Sig>
  struct mof_internal;
  template<class R, class...Args>
  struct mof_internal {
    virtual ~mof_internal() {};
    // 4 overloads, because I'm insane:
    virtual R invoke( Args&&... args ) const& = 0;
    virtual R invoke( Args&&... args ) & = 0;
    virtual R invoke( Args&&... args ) const&& = 0;
    virtual R invoke( Args&&... args ) && = 0;
  };

  template<class F, class Sig>
  struct mof_pimpl;
  template<class R, class...Args, class F>
  struct mof_pimpl<F, R(Args...)>:mof_internal<R(Args...)> {
    F f;
    virtual R invoke( Args&&... args ) const&  override { return f( std::forward<Args>(args)... ); }
    virtual R invoke( Args&&... args )      &  override { return f( std::forward<Args>(args)... ); }
    virtual R invoke( Args&&... args ) const&& override { return std::move(f)( std::forward<Args>(args)... ); }
    virtual R invoke( Args&&... args )      && override { return std::move(f)( std::forward<Args>(args)... ); }
  };
}

template<class R, class...Args>
struct move_only_function<R(Args)> {
  move_only_function(move_only_function const&)=delete;
  move_only_function(move_only_function &&)=default;
  move_only_function(std::nullptr_t):move_only_function() {}
  move_only_function() = default;
  explicit operator bool() const { return pImpl; }
  bool operator!() const { return !*this; }
  R operator()(Args...args)     & { return                  pImpl().invoke(std::forward<Args>(args)...); }
  R operator()(Args...args)const& { return                  pImpl().invoke(std::forward<Args>(args)...); }
  R operator()(Args...args)     &&{ return std::move(*this).pImpl().invoke(std::forward<Args>(args)...); }
  R operator()(Args...args)const&&{ return std::move(*this).pImpl().invoke(std::forward<Args>(args)...); }

  template<class F,class=std::enable_if_t<can_invoke<decay_t<F>(Args...)>>
  move_only_function(F&& f):
    m_pImpl( std::make_unique<details::mof_pimpl<std::decay_t<F>, R(Args...)>>( std::forward<F>(f) ) )
  {}
private:
  using internal = details::mof_internal<R(Args...)>;
  std::unique_ptr<internal> m_pImpl;

  // rvalue helpers:
  internal      &  pImpl()      &  { return *m_pImpl.get(); }
  internal const&  pImpl() const&  { return *m_pImpl.get(); }
  internal      && pImpl()      && { return std::move(*m_pImpl.get()); }
  internal const&& pImpl() const&& { return std::move(*m_pImpl.get()); } // mostly useless
 };

not tested, just spewed the code. The can_invoke gives the constructor basic SFINAE -- you can add "return type converts properly" and "void return type means we ignore the return" if you like.

Now we rework your code. First, your task are move-only functions, not functions:

std::vector<move_only_function<X>> mTasks;

Next, we store the R type calculation once, and use it again:

template<class F, class... Args, class R=std::result_of_t<std::decay<F>_&&(std::decay_t<Args>&&...)>>
std::future<R>
push(F&& f, Args&&... args)
{
  auto tuple_args=std::make_tuple(std::forward<Args>(args)...)];

  // lambda will only be called once:
  std::packaged_task<R()> task([f=std::forward<F>(f),args=std::move(tuple_args)]
    return invoke_tuple( std::move(f), std::move(args) );
  });

   auto future = func.get_future();

  // for some reason I get a compilation error in clang if I get rid of the `=, ` in this capture:
  mTasks.emplace_back( std::move(task) );

  return future;
}

we stuff the arguments into a tuple, pass that tuple into a lambda, and invoke the tuple in a "only do this once" kind of way within the lambda. As we will only invoke the function once, we optimize the lambda for that case.

A packaged_task<R()> is compatible with a move_only_function<R()> unlike a std::function<R()>, so we can just move it into our vector. The std::future we get from it should work fine even though we got it before the move.

This should reduce your overhead by a bit. Of course, there is lots of boilerplate.

I have not compiled any of the above code, I just spewed it out, so the odds it all compiles are low. But the errors should mostly be tpyos.

Randomly, I decided to give move_only_function 4 different () overloads (rvalue/lvalue and const/not). I could have added volatile, but that seems reckless. Which increase boilerplate, admittedly.

Also my move_only_function lacks the "get at the underlying stored stuff" operation that std::function has. Feel free to type erase that if you like. And it treats (R(*)(Args...))0 as if it was a real function pointer (I return true when cast to bool, not like null: type erasure of convert-to-bool might be worthwhile for a more industrial quality implementation.

I rewrote std::function because std lacks a std::move_only_function, and the concept in general is a useful one (as evidenced by packaged_task). Your solution makes your callable movable by wrapping it with a std::shared_ptr.

If you don't like the above boilerplate, consider writing make_copyable(F&&), which takes an function object F and wraps it up using your shared_ptr technique to make it copyable. You can even add SFINAE to avoid doing it if it is already copyable (and call it ensure_copyable).

Then your original code would be cleaner, as you'd just make the packaged_task copyable, then store that.

template<class F>
auto make_function_copyable( F&& f ) {
  auto sp = std::make_shared<std::decay_t<F>>(std::forward<F>(f));
  return [sp](auto&&...args){return (*sp)(std::forward<decltype(args)>(args)...); }
}
template<class F, class... Args, class R=std::result_of_t<std::decay<F>_&&(std::decay_t<Args>&&...)>>
std::future<R>
push(F&& f, Args&&... args)
{
  auto tuple_args=std::make_tuple(std::forward<Args>(args)...)];

  // lambda will only be called once:
  std::packaged_task<R()> task([f=std::forward<F>(f),args=std::move(tuple_args)]
    return invoke_tuple( std::move(f), std::move(args) );
  });

  auto future = func.get_future();

  // for some reason I get a compilation error in clang if I get rid of the `=, ` in this capture:
  mTasks.emplace_back( make_function_copyable( std::move(task) ) );

  return future;
}

this still requires the invoke_tuple boilerplate above, mainly because I dislike bind.