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53 lines
2.9 KiB
53 lines
2.9 KiB
5 years ago
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From 64d0115dcda445a5b1c069f696a363730f654425 Mon Sep 17 00:00:00 2001
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From: David Tardon <dtardon@redhat.com>
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Date: Thu, 2 May 2019 12:55:04 +0200
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Subject: [PATCH] avoid possible hang if our child process hangs
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If there is one or more unexpected child processes that terminate, but
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the "main" child process hangs, we will loop through the terminated
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children and then, eventually, get stuck in the waitpid() call. Let's
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repeat the main cycle if that situation happens, as that allows us to
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finish after timeout.
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Related: #1697909
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---
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src/udev/udev-event.c | 11 +++++++++--
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1 file changed, 9 insertions(+), 2 deletions(-)
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diff --git a/src/udev/udev-event.c b/src/udev/udev-event.c
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index 5550ec93de..79b8614ec2 100644
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--- a/src/udev/udev-event.c
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+++ b/src/udev/udev-event.c
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@@ -605,6 +605,7 @@ static int spawn_wait(struct udev_event *event,
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struct signalfd_siginfo fdsi;
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int status, r;
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ssize_t size;
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+ bool wait_again = false;
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size = read(event->fd_signal, &fdsi, sizeof(struct signalfd_siginfo));
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if (size != sizeof(struct signalfd_siginfo))
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@@ -622,15 +623,21 @@ static int spawn_wait(struct udev_event *event,
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In case the PID we wait for also exited the kernel could coalesce SIGCHLDs and we won't get second SIGCHLD
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on the signalfd. We can't know if coalescing happened or not, hence we need to call waitpid() in a loop until
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the PID we care for exits, we if it haven't already. */
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- while (child_exited < 0) {
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- r = waitpid(-1, &status, 0);
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+ while (child_exited < 0 && !wait_again) {
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+ r = waitpid(-1, &status, WNOHANG);
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if (r < 0 && errno == EINTR)
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continue;
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else if (r < 0)
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break;
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else if (r == pid)
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child_exited = 0;
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+ else if (r == 0)
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+ wait_again = true;
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}
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+
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+ /* Only unexpected process(es) have been terminated. Continue waiting */
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+ if (wait_again)
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+ continue;
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}
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/* We didn't wait for child yet, let's do that now */
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