[MESOS-4611] Passing a lambda to dispatch() always matches the template returning void

• Type: Bug
• Status: Resolved
• Priority: Major
• Resolution: Fixed
• Fix Version/s: 1.0.0
• Component/s: libprocess
• Labels: dispatch libprocess mesosphere
• Sprint: Mesosphere Sprint 38
• Story Points: 5
• Assignee: haosdent
• Reporter: Kevin Klues
• Shepherd: Michael Park
• Created: 05/Feb/16 20:55
• Updated: 26/Nov/18 12:21
• Resolved: 26/Nov/18 12:21

Description

The following idiom does not currently compile:

  Future<Nothing> initialized = dispatch(pid, [] () -> Nothing {
    return Nothing();
  });

This seems non-intuitive because the following template exists for dispatch:

template <typename R>
Future<R> dispatch(const UPID& pid, const std::function<R()>& f)
{
  std::shared_ptr<Promise<R>> promise(new Promise<R>()); 
 
  std::shared_ptr<std::function<void(ProcessBase*)>> f_(
      new std::function<void(ProcessBase*)>(
          [=](ProcessBase*) {
            promise->set(f());
          }));

  internal::dispatch(pid, f_);
  
  return promise->future();
}     

However, lambdas cannot be implicitly cast to a corresponding std::function<R()> type.
To make this work, you have to explicitly type the lambda before passing it to dispatch.

  std::function<Nothing()> f = []() { return Nothing(); };
  Future<Nothing> initialized = dispatch(pid, f);

We should add template support to allow lambdas to be passed to dispatch() without explicit typing.