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3843153
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0
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0
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2010-10-01T21:26:55.163
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2010-10-03T17:14:15.863
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2010-10-03T17:14:15.863
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444705
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444705
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3842410
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2
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First, generators are probably a red herring; don't worry about them. The canonical way to solve this kind producer-consumer problem in Python is using the built-in <a href="http://docs.python.org/library/queue.html" rel="nofollow"><code>queue</code></a> module. It acts as an intermediary, allowing your producer thread to keep grabbing/processing data from the kernel into the queue, and your consumer thread to send queue data to the server, without their respective blocking I/O calls interfering with one another. Here's a sketch of the basic idea, without the details filled in: <pre><code>from queue import Queue class Application(object): def __init__(self): self.q = Queue() self.running = False # From kernel to queue def produce(self): while self.running: data = read_from_kernel() self.q.put(data) # From queue to server def consume(self): while self.running: data = self.q.get() send_to_server(data) # Start producer thread, then consume def run(): try: self.running = True producer = Thread(target=self.produce) producer.start() self.consume() finally: self.running = False </code></pre> If <code>self.running</code> is set to False, the above code's <code>produce</code> method will still block inside the <code>read_from_kernel</code> until its next return before exiting itself, but there's little Python can do about that. Whatever system call you use must support this somehow: if it's an actual <code>read</code>, for example, your options would include: <ul> <li>A short timeout, plus retry handling</li> <li>Non-blocking I/O (but in this case you might want to investigate a framework based around this, like <a href="http://twistedmatrix.com/" rel="nofollow">Twisted Python</a>)</li> </ul>
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1. POIs there a way to cleanly exit out of a thread which is processing data from a (never-ending) generator?
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