>
> +static void notify_process(pid_t pid, u64 fault_addr)
> +{
> + int rc;
> + struct kernel_siginfo info;
> +
> + memset(&info, 0, sizeof(info));
> +
> + info.si_signo = SIGSEGV;
> + info.si_errno = EFAULT;
> + info.si_code = SEGV_MAPERR;
> +
> + info.si_addr = (void *)fault_addr;
> + rcu_read_lock();
> + rc = kill_pid_info(SIGSEGV, &info, find_vpid(pid));
> + rcu_read_unlock();
> +
> + pr_devel("%s(): pid %d kill_proc_info() rc %d\n", __func__, pid, rc);
> +}
Shouldn't this use force_sig_fault_to_task instead?
> + /*
> + * User space passed invalid CSB address, Notify process with
> + * SEGV signal.
> + */
> + tsk = get_pid_task(window->pid, PIDTYPE_PID);
> + /*
> + * Send window will be closed after processing all NX requests
> + * and process exits after closing all windows. In multi-thread
> + * applications, thread may not exists, but does not close FD
> + * (means send window) upon exit. Parent thread (tgid) can use
> + * and close the window later.
> + */
> + if (tsk) {
> + if (tsk->flags & PF_EXITING)
> + task_exit = 1;
> + put_task_struct(tsk);
> + pid = vas_window_pid(window);
The pid is later used for sending the signal again, why not keep the
reference?
> + } else {
> + pid = vas_window_tgid(window);
> +
> + rcu_read_lock();
> + tsk = find_task_by_vpid(pid);
> + if (!tsk) {
> + rcu_read_unlock();
> + return;
> + }
> + if (tsk->flags & PF_EXITING)
> + task_exit = 1;
> + rcu_read_unlock();
Why does this not need a reference to the task, but the other one does?
