On 15/07/2013 04:04, Steven D'Aprano wrote:
On Mon, 15 Jul 2013 10:27:45 +0800, Gildor Oronar wrote:A currency exchange thread updates exchange rate once a minute. If the thread faield to update currency rate for 5 hours, it should inform main() for a clean exit. This has to be done gracefully, because main() could be doing something delicate. I, a newbie, read all the thread sync tool, and wasn't sure which one to use. In fact I am not sure if there is a need of thread sync, because there is no racing cond. I thought of this naive way: class CurrencyExchange(): def __init__(in_case_callback): this.callback = in_case_callbackYou need to declare the instance parameter, which is conventionally called "self" not "this". Also, your class needs to inherit from Thread, and critically it MUST call the superclass __init__. So: class CurrencyExchange(threading.Thread): def __init__(self, in_case_callback): super(CurrencyExchange, self).__init__() self.callback = in_case_callback But I'm not sure that a callback is the right approach here. See below.def __run__():Likewise, you need a "self" parameter.while time.time() - self.rate_timestamp < 5*3600: ... # update exchange rate if success:
The "==" in this line should, of course, be "=":
self.rate_timestamp == time.time() time.sleep(60) this.callback() # rate not updated 5 hours, a crisisI think that a cleaner way is to just set a flag on the thread instance. Initiate it with: self.updates_seen = True in the __init__ method, and then add this after the while loop: self.updates_seen = Falsedef main(): def callback() Go_On = FalseI don't believe this callback will work, because it will simply create a local variable call "Go_On", not change the non-local variable. In Python 3, you can use the nonlocal keyword to get what you want, but I think a better approach is with a flag on the thread.agio = CurrencyExchange(in_case = callback) agio.start() Go_On = True while Go_On: do_something_delicate(rate_supplied_by=agio)Change to: while agio.updates_seen: do_something_delicate...
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