In looking over the kernel for preemption hazards I found the following:
The following function is from select.c. Can someone help me to
understand why do_pollfd() is being called with the task state set to
TASK_INTERRUPTIBLE? do_pollfd() calls the driver poll routine. Is it
expected that the driver might in some way call schedule() and put the
current task to sleep? Or would this be a big NO-NO. I think the state
should be set just prior to the schedule_timeout() call. Would this
impair some functionality? The reason I am asking is, if the kernel is
preempted between the two lines in question, the it appears that the
task will be put to sleep, _forever_! And yes I know the kernel is not
preemptable..._yet_.
If I am correct, then moving the line would also eliminate the need to
set the state back to TASK_RUNNING.
George
static int do_poll(unsigned int nfds, unsigned int nchunks, unsigned int
nleft,
struct pollfd *fds[], poll_table *wait, long timeout)
{
int count = 0;
for (;;) {
unsigned int i;
set_current_state(TASK_INTERRUPTIBLE);
for (i=0; i < nchunks; i++)
do_pollfd(POLLFD_PER_PAGE, fds[i], &wait, &count);
if (nleft)
do_pollfd(nleft, fds[nchunks], &wait, &count);
wait = NULL;
if (count || !timeout || signal_pending(current))
break;
timeout = schedule_timeout(timeout);
}
current->state = TASK_RUNNING;
return count;
}
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