[Twisted-Python] how to get the result of a callLater-scheduled func?

[Stefan Rank]

Hi,

I am scheduling a function with `callLater`.
The function happens to return a Deferred (but it might return anything),
and I was searching for an easy way to add a callback in order to
get the result.
I know a workaround by creating a Deferred first, adding the call as 
callback and another callback to get the result, but this seems overly 
complicated (see below).
Am I missing something?

So I set up some function to be scheduled (happens to be an 
inlineCallbacks-generator for testing, but I think this should not be 
relevant here)::

     >>> from twisted.internet import defer, reactor
     >>> @defer.inlineCallbacks
     ... def test(name):
     ...     print name, 'start'
     ...     yield # yielding a non-Deferred does nothing
     ...     print name, 'end'
     ...     yield defer.returnValue('uiuiui')

I know this works::

     >>> deftest = defer.Deferred().addCallback(test)
     >>> jamesdelcall = reactor.callLater(0, deftest.callback, 'James')
     >>> def stopverbose(whatever):
     ...     print 'stopping, result:', whatever
     ...     reactor.stop()
     >>> deftest.addCallback(stopverbose)
     >>> reactor.run()
     James start
     James end
     stopping, result: uiuiui

What I expected to be able to do::

     >>> jamesdelcall = reactor.callLater(0, test, 'James')
     >>> def stopverbose(whatever):
     ...     print 'stopping, result:', whatever
     ...     reactor.stop()
     >>> jamesdelcal.gethisdeferred.addCallback(stopverbose) # <- MAGIC
     >>> reactor.run()
     James start
     James end
     stopping, result: uiuiui

I looked at the code of `ReactorBase.callLater` and 
`ReactorBase.runUntilCurrent`:
Would it be possible to add a `Deferred` to `DelayedCall` (possibly 
created lazily, or maybe even make DelayedCall a subclass of Deferred),
so that such code is possible, or would this somehow destroy the 
mainloop logic?

cheers,
stefan