bpo-45274: Fix Thread._wait_for_tstate_lock() race condition (GH-28532) (GH-28580)

Fix a race condition in the Thread.join() method of the threading
module. If the function is interrupted by a signal and the signal
handler raises an exception, make sure that the thread remains in a
consistent state to prevent a deadlock.
(cherry picked from commit a22be4943c)

Co-authored-by: Victor Stinner <vstinner@python.org>

Lib/threading.py

@@ -1100,11 +1100,24 @@ class Thread:
         # If the lock is acquired, the C code is done, and self._stop() is
         # called.  That sets ._is_stopped to True, and ._tstate_lock to None.
         lock = self._tstate_lock
-        if lock is None:  # already determined that the C code is done
+        if lock is None:
+            # already determined that the C code is done
             assert self._is_stopped
-        elif lock.acquire(block, timeout):
+            return
-            lock.release()
+
-            self._stop()
+        try:
+            if lock.acquire(block, timeout):
+                lock.release()
+                self._stop()
+        except:
+            if lock.locked():
+                # bpo-45274: lock.acquire() acquired the lock, but the function
+                # was interrupted with an exception before reaching the
+                # lock.release(). It can happen if a signal handler raises an
+                # exception, like CTRL+C which raises KeyboardInterrupt.
+                lock.release()
+                self._stop()
+            raise
 
     @property
     def name(self):

Misc/NEWS.d/next/Library/2021-09-23-22-17-26.bpo-45274.gPpa4E.rst

@@ -0,0 +1,5 @@
+Fix a race condition in the :meth:`Thread.join() <threading.Thread.join>`
+method of the :mod:`threading` module. If the function is interrupted by a
+signal and the signal handler raises an exception, make sure that the thread
+remains in a consistent state to prevent a deadlock. Patch by Victor
+Stinner.