On Thursday, 6 February 2014 at 16:07:51 UTC, Andrea Fontana wrote:
> On Thursday, 6 February 2014 at 14:52:36 UTC, Cooler wrote:
>> On Thursday, 6 February 2014 at 14:42:57 UTC, Cooler wrote:
>>> On Thursday, 6 February 2014 at 11:30:17 UTC, Andrea Fontana wrote:
>>>> On Wednesday, 5 February 2014 at 15:38:14 UTC, Cooler wrote:
>>>>> On Tuesday, 4 February 2014 at 03:26:04 UTC, Dan Killebrew wrote:
>>>>>>>> It seems to me that worker threads will continue as long as the queue isn't empty. So if a task adds another task to the pool, some worker will process the newly enqueued task.
>>>>>>>
>>>>>>> No. After taskPool.finish() no way to add new tasks to the queue. taskPool.put will not add new tasks.
>>>>>>
>>>>>> Then perhaps you need to create a new TaskPool (and make sure that workers add their tasks to the correct task pool), so that you can wait on the first task pool, then wait on the second task pool, etc.
>>>>>>
>>>>>> auto phase1 = new TaskPool();
>>>>>> //make sure all new tasks are added to phase1
>>>>>> phase1.finish(true);
>>>>>>
>>>>>> auto phase2 = new TaskPool();
>>>>>> //make sure all new tasks are added to phase2
>>>>>> phase2.finish(true);
>>>>>
>>>>> Will not help. I don't know beforehand what tasks will be
>>>>> created. procData is recursive and it decides create new task or
>>>>> not.
>>>>
>>>>
>>>> Something like this? (not tested...)
>>>>
>>>> shared bool more = true;
>>>> ...
>>>> ...
>>>> ...
>>>>
>>>> void procData(){
>>>> if(...)
>>>> {
>>>>  taskPool.put(task(&procData));
>>>>  more = true;
>>>> }
>>>> }
>>>>
>>>> while(true)
>>>> {
>>>> taskPool.finish(true);
>>>> if (!more) break;
>>>> else more = false;
>>>> }
>>>
>>> It is closer, but after taskPool.finish() all tries to taskPool.put() will be rejected. Let's me clear example.
>>>
>>> import std.stdio, std.parallelism, core.thread;
>>>
>>> shared int i;
>>>
>>> void procData(){
>>> synchronized ++i;
>>> if(i >= 100)
>>>   return;
>>> foreach(i; 0 .. 100)
>>>   taskPool.put(task(&procData)); // New tasks will be rejected after
>>>                                  // taskPool.finish()
>>> }
>>>
>>> void main(){
>>> taskPool.put(task(&procData));
>>> Thread.sleep(1.msecs); // The final output of "i" depends on duration here
>>> taskPool.finish(true);
>>> writefln("i = %s", i);
>>> }
>>>
>>> In the example above the total number of tasks executed depends on sleep duration.
>>
>> Forgot to say - I know how to solve the topic problem. My
>> question is "What is the BEST way?".
>> One of my idea - may be introduce new function, named for example
>> "wait", that will block until there are working tasks?
>
> What about sync ++taskCount when you put() something and --taskCount when task is done? And on main while(i > 0) Thread.yield(); ?

Something like this:

import std.stdio, std.parallelism, core.thread;
import std.random;

shared size_t taskCount;
shared size_t i;

void procData()
in  { synchronized ++i; }
out { synchronized --taskCount; }
body
{
    if (i > 100)
        return;

    foreach(i; 0 .. 100)
    {
        taskPool.put(task(&procData));
        synchronized ++taskCount;
    }

}

void main(){

    taskCount = 2;
    taskPool.put(task(&procData));
    taskPool.put(task(&procData));

    while(taskCount > 0)
        Thread.yield();
}

On Mon, 2014-02-03 at 00:00 +0000, Cooler wrote:
> I have several tasks. Each task may or may not create another task. What is the best way to wait until all tasks finished?

What you are describing here is a classic fork/join architecture. The tasks are structured as a tree with synchronization handled by the sub-nodes.

As far as I am aware std.parallelism focuses on data parallelism which is a scatter/gather (aka map/reduce) model of just a single layer.

All the code fragments in the thread have, I believe, been predicated on working with a thread pool as an explicit global entity. I think the problems have stemmed from taking this viewpoint.

I would suggest following the way the Java fork/join framework (based on Doug Lea's original) works. There is an underlying global thread pool, but the user code uses the fork/join abstraction layer in order to create the tree of synchronization dependencies. In this case instead of working with tasks directly there needs to be a type whose job it is to be a non-leaf node in the tree that handles synchronization whilst nonetheless creating tasks and submitting them to the pool.

This is clearly something that could turn into an addition to std.parallelism or be std.forkjoin.

Sorry I have no actual code to offer, but the overall design of what is needed is well understood, at least in the Java context. C++ has a long way to go to catch up, as does D.

The other thing that then sits on this is lazy stream parallelism, which is what Java 8 is adding to the mix.

-- 
Russel. ============================================================================= Dr Russel Winder      t: +44 20 7585 2200   voip: sip:russel.winder@ekiga.net 41 Buckmaster Road    m: +44 7770 465 077   xmpp: russel@winder.org.uk London SW11 1EN, UK   w: www.russel.org.uk  skype: russel_winder

On Thursday, 6 February 2014 at 16:19:38 UTC, Andrea Fontana
wrote:
> On Thursday, 6 February 2014 at 16:07:51 UTC, Andrea Fontana wrote:
>> On Thursday, 6 February 2014 at 14:52:36 UTC, Cooler wrote:
>>> On Thursday, 6 February 2014 at 14:42:57 UTC, Cooler wrote:
>>>> On Thursday, 6 February 2014 at 11:30:17 UTC, Andrea Fontana wrote:
>>>>> On Wednesday, 5 February 2014 at 15:38:14 UTC, Cooler wrote:
>>>>>> On Tuesday, 4 February 2014 at 03:26:04 UTC, Dan Killebrew wrote:
>>>>>>>>> It seems to me that worker threads will continue as long as the queue isn't empty. So if a task adds another task to the pool, some worker will process the newly enqueued task.
>>>>>>>>
>>>>>>>> No. After taskPool.finish() no way to add new tasks to the queue. taskPool.put will not add new tasks.
>>>>>>>
>>>>>>> Then perhaps you need to create a new TaskPool (and make sure that workers add their tasks to the correct task pool), so that you can wait on the first task pool, then wait on the second task pool, etc.
>>>>>>>
>>>>>>> auto phase1 = new TaskPool();
>>>>>>> //make sure all new tasks are added to phase1
>>>>>>> phase1.finish(true);
>>>>>>>
>>>>>>> auto phase2 = new TaskPool();
>>>>>>> //make sure all new tasks are added to phase2
>>>>>>> phase2.finish(true);
>>>>>>
>>>>>> Will not help. I don't know beforehand what tasks will be
>>>>>> created. procData is recursive and it decides create new task or
>>>>>> not.
>>>>>
>>>>>
>>>>> Something like this? (not tested...)
>>>>>
>>>>> shared bool more = true;
>>>>> ...
>>>>> ...
>>>>> ...
>>>>>
>>>>> void procData(){
>>>>> if(...)
>>>>> {
>>>>> taskPool.put(task(&procData));
>>>>> more = true;
>>>>> }
>>>>> }
>>>>>
>>>>> while(true)
>>>>> {
>>>>> taskPool.finish(true);
>>>>> if (!more) break;
>>>>> else more = false;
>>>>> }
>>>>
>>>> It is closer, but after taskPool.finish() all tries to taskPool.put() will be rejected. Let's me clear example.
>>>>
>>>> import std.stdio, std.parallelism, core.thread;
>>>>
>>>> shared int i;
>>>>
>>>> void procData(){
>>>> synchronized ++i;
>>>> if(i >= 100)
>>>>  return;
>>>> foreach(i; 0 .. 100)
>>>>  taskPool.put(task(&procData)); // New tasks will be rejected after
>>>>                                 // taskPool.finish()
>>>> }
>>>>
>>>> void main(){
>>>> taskPool.put(task(&procData));
>>>> Thread.sleep(1.msecs); // The final output of "i" depends on duration here
>>>> taskPool.finish(true);
>>>> writefln("i = %s", i);
>>>> }
>>>>
>>>> In the example above the total number of tasks executed depends on sleep duration.
>>>
>>> Forgot to say - I know how to solve the topic problem. My
>>> question is "What is the BEST way?".
>>> One of my idea - may be introduce new function, named for example
>>> "wait", that will block until there are working tasks?
>>
>> What about sync ++taskCount when you put() something and --taskCount when task is done? And on main while(i > 0) Thread.yield(); ?
>
> Something like this:
>
> import std.stdio, std.parallelism, core.thread;
> import std.random;
>
> shared size_t taskCount;
> shared size_t i;
>
> void procData()
> in  { synchronized ++i; }
> out { synchronized --taskCount; }
> body
> {
>     if (i > 100)
>         return;
>
>     foreach(i; 0 .. 100)
>     {
>         taskPool.put(task(&procData));
>         synchronized ++taskCount;
>     }
>
> }
>
> void main(){
>
>     taskCount = 2;
>     taskPool.put(task(&procData));
>     taskPool.put(task(&procData));
>
>     while(taskCount > 0)
>         Thread.yield();
> }

Now I do almost the same in my program, but instead of while(...)
Thread.yield() I wait on semaphore to be notified. And in threads
when --taskCount reaches 0, i do Semaphore.notify(). But all of
this don't look beautiful enough. As I mention above - may be
introduce new function, named for example "wait", that will block
TaskPool until there are working tasks? If such situation
encounters frequently, may be it is worth to add Phobos some
functionality?

On Thursday, 6 February 2014 at 17:58:47 UTC, Russel Winder wrote:
> On Mon, 2014-02-03 at 00:00 +0000, Cooler wrote:
>> I have several tasks. Each task may or may not create another task. What is the best way to wait until all tasks finished?
>
> What you are describing here is a classic fork/join architecture. The
> tasks are structured as a tree with synchronization handled by the
> sub-nodes.
>
> As far as I am aware std.parallelism focuses on data parallelism which
> is a scatter/gather (aka map/reduce) model of just a single layer.
>
> All the code fragments in the thread have, I believe, been predicated on
> working with a thread pool as an explicit global entity. I think the
> problems have stemmed from taking this viewpoint.
>
> I would suggest following the way the Java fork/join framework (based on
> Doug Lea's original) works. There is an underlying global thread pool,
> but the user code uses the fork/join abstraction layer in order to
> create the tree of synchronization dependencies. In this case instead of
> working with tasks directly there needs to be a type whose job it is to
> be a non-leaf node in the tree that handles synchronization whilst
> nonetheless creating tasks and submitting them to the pool.
>
> This is clearly something that could turn into an addition to
> std.parallelism or be std.forkjoin.
>
> Sorry I have no actual code to offer, but the overall design of what is
> needed is well understood, at least in the Java context. C++ has a long
> way to go to catch up, as does D.
>
> The other thing that then sits on this is lazy stream parallelism, which
> is what Java 8 is adding to the mix.

Thank you for your verbose answer. I think the solution could be shorter and simpler :)