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nodes-api: enforce single thread for the processor
On hyperthreaded CPUs we get two threads operating on the queue by default on each core. This cases two threads to process queue items. This results in pytorch errors and sometimes generates garbage. Locking this to single thread makes sense because we are bound by the number of GPUs in the system, not by CPU cores. And to parallelize across GPUs we should just start multiple processors (and use async instead of threading) Fixes #3289
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@ -1,5 +1,5 @@
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import traceback
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from threading import Event, Thread
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from threading import Event, Thread, BoundedSemaphore
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from ..invocations.baseinvocation import InvocationContext
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from .invocation_queue import InvocationQueueItem
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@ -10,8 +10,11 @@ class DefaultInvocationProcessor(InvocationProcessorABC):
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__invoker_thread: Thread
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__stop_event: Event
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__invoker: Invoker
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__threadLimit: BoundedSemaphore
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def start(self, invoker) -> None:
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# if we do want multithreading at some point, we could make this configurable
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self.__threadLimit = BoundedSemaphore(1)
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self.__invoker = invoker
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self.__stop_event = Event()
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self.__invoker_thread = Thread(
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@ -20,7 +23,7 @@ class DefaultInvocationProcessor(InvocationProcessorABC):
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kwargs=dict(stop_event=self.__stop_event),
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)
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self.__invoker_thread.daemon = (
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True # TODO: probably better to just not use threads?
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True # TODO: make async and do not use threads
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)
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self.__invoker_thread.start()
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@ -29,6 +32,7 @@ class DefaultInvocationProcessor(InvocationProcessorABC):
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def __process(self, stop_event: Event):
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try:
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self.__threadLimit.acquire()
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while not stop_event.is_set():
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queue_item: InvocationQueueItem = self.__invoker.services.queue.get()
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if not queue_item: # Probably stopping
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@ -110,7 +114,7 @@ class DefaultInvocationProcessor(InvocationProcessorABC):
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)
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pass
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# Check queue to see if this is canceled, and skip if so
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if self.__invoker.services.queue.is_canceled(
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graph_execution_state.id
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@ -127,4 +131,6 @@ class DefaultInvocationProcessor(InvocationProcessorABC):
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)
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except KeyboardInterrupt:
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... # Log something?
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pass # Log something? KeyboardInterrupt is probably not going to be seen by the processor
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finally:
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self.__threadLimit.release()
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