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_callback
You 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:
> self.rate_timestamp == time.time()
> time.sleep(60)
> this.callback() # rate not updated 5 hours, a crisis
I 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 = False
> def main():
> def callback()
> Go_On = False
I 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...
--
Steven
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