with timeout(...):

[Jean-Paul Calderone]

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?

Jean-Paul