Race condition in Search.ozf

[Luis Quesada]

Harmon Nine wrote:

>So the program would look something like this?
>
>==========================
>Sobj = {New Search.object script(Space OptProc)}
>
>proc {WillContinue NextSolution} {SObj next(NextSolution)} end
>proc {WillStop NextSolution} NextSolution = stopped end
>
>P = {NewCell WillContinue}
>
>Thread 1:
>proc {GetNextSolution}
>  NextSolution
> in
>  {@P NextSolution}
>
>  if NextSolution == stopped then
>    skip
>  else
>    Process NextSolution
>    {GetNextSolution}
>  end
>
>end
>
>Thread 2:
>proc {StopThread}
>  P := WillStop
>  {Sobj stop}
>end
>==========================
>
>  
>
What I mean is to redifine the Order that is passed as a paramenter to 
the script method:

P={NewCell "your order procedure"}
Order= proc{$ Old New} {@P Old New} end
Sobj={New Search.object script(Script Order)}

Thread 1:
proc {GetNextSolution}

  NextSolution
 in
  {Sobj next(NextSolution)}

  if {Member NextSolution [stopped nil]} then
    skip
  else
    Process NextSolution
    {GetNextSolution}
  end

end

Thread 2:
proc {Stop}
   P:=proc {$ _ _} fail end
   {Sobj stop}
end


>This is a better solution, i.e. it is deterministic in that the call in
>Thread 2 to the StopThread proc is guaranteed to (eventually) stop
>calculation.  However there's still a race that has a (very small)
>chance of causing the calculations not to stop immediately when the
>"stop" method is called on SObj.  An immediate termination would be
>necessary on receipt of a user input event, or the expiration of a
>soft-real-time timer.
>
>Scenario:
>@P has value WillContinue
>In Thread 1 (top of GetNextSolution proc), @P is dereferenced and thus
>WillContinue is called.
>Program execution in Thread 1 is now in the WillContinue proc, but
>Thread 1 is pre-empted *before* the {SObj next(NextSolution}} call.
>
>Thread 2 fires up on user input event, changes the value of P to
>WillStop and calls the stop method on SObj.
>This call to the stop method sets the 'isStopped' flag for the Search
>object SObj.
>Thread 2 is pre-empted.
>  
>

If Stop has been executed before {Sobj next(NextSolution)}, 
{Sobj next(NextSolution)} will return immediately because of the failure exception.

Luis