On Thu, 29 Mar 2007 11:30:04 -0500, Nick Craig-Wood <[EMAIL PROTECTED]> wrote: >Diez B. Roggisch <[EMAIL PROTECTED]> 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 -- http://mail.python.org/mailman/listinfo/python-list
