exarkun at divmod.com
Thu Mar 29 19:15:01 CEST 2007
On Thu, 29 Mar 2007 11:30:04 -0500, Nick Craig-Wood <nick at craig-wood.com> wrote:
>Diez B. Roggisch <deets at nospam.web.de> wrote:
>> > I beleive the convention is when calling an OS function which might
>> > block the global interpreter lock is dropped, thus allowing other
>> > python bytecode to run.
>> So what? That doesn't help you, as you are single-threaded here. The
>> released lock won't prevent the called C-code from taking as long as it
>> wants. |And there is nothing you can do about that.
>I'm assuming that the timeout function is running in a thread...
What does it do when the timeout expires? How does it interrupt recv(2)
or write(2) or `for (int i = 0; i < (unsigned)-1; ++i);'?
This is what we're talking about, right?
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