BUG/MEDIUM: wdt: Don't ignore WDTSIG and DEBUGSIG in __signal_process_queue().
When running __signal_process_queue(), we ignore most signals. We can't,
however, ignore WDTSIG and DEBUGSIG, otherwise that thread may end up
waiting for another one that could hold a glibc lock, while the other thread
wait for this one to enter debug_handler().
So make sure WDTSIG and DEBUGSIG aren't ignored, if they are defined.
This probably explains the watchdog deadlock described in github issue
This should be backported to 2.1, 2.0 and 1.9.
(cherry picked from commit b0198cc4132381910cdeb9b5a867632b8b83262c)
Signed-off-by: Willy Tarreau <w@1wt.eu>
(cherry picked from commit 565ed0dc0465ad67891b4789f563db96c3143365)
Signed-off-by: Willy Tarreau <w@1wt.eu>
diff --git a/src/signal.c b/src/signal.c
index 20236fa..288ef00 100644
--- a/src/signal.c
+++ b/src/signal.c
@@ -114,11 +114,19 @@
/* man sigprocmask: If SIGBUS, SIGFPE, SIGILL, or SIGSEGV are
generated while they are blocked, the result is undefined, unless
the signal was generated by kill(2),
- sigqueue(3), or raise(3) */
+ sigqueue(3), or raise(3).
+ Do not ignore WDTSIG or DEBUGSIG either, or it may deadlock the
+ watchdog */
sigdelset(&blocked_sig, SIGBUS);
sigdelset(&blocked_sig, SIGFPE);
sigdelset(&blocked_sig, SIGILL);
sigdelset(&blocked_sig, SIGSEGV);
+#ifdef DEBUGSIG
+ sigdelset(&blocked_sig, DEBUGSIG);
+#endif
+#ifdef WDTSIG
+ sigdelset(&blocked_sig, WDTSIG);
+#endif
for (sig = 0; sig < MAX_SIGNAL; sig++)
LIST_INIT(&signal_state[sig].handlers);
}