Messages in this thread | | | Date | Mon, 17 Jul 2000 08:21:13 -0700 | From | George Anzinger <> | Subject | Looking at select, a problem or functionality? |
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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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