Commit c5abd63e authored by Kirill Smelkov's avatar Kirill Smelkov Committed by Victor Stinner

bpo-38106: Fix race in pthread PyThread_release_lock() (GH-16047)

Fix race in PyThread_release_lock that was leading to memory corruption and
deadlocks. The fix applies to POSIX systems where Python locks are implemented
with mutex and condition variable because POSIX semaphores are either not
provided, or are known to be broken. One particular example of such system is
macOS.

On Darwin, even though this is considered as POSIX, Python uses
mutex+condition variable to implement its lock, and, as of 2019-08-28, Py2.7
implementation, even though similar issue was fixed for Py3 in 2012, contains
synchronization bug: the condition is signalled after mutex unlock while the
correct protocol is to signal condition from under mutex:

  https://github.com/python/cpython/blob/v2.7.16-127-g0229b56d8c0/Python/thread_pthread.h#L486-L506
  https://github.com/python/cpython/commit/187aa545165d (py3 fix)

PyPy has the same bug for both pypy2 and pypy3:

  https://bitbucket.org/pypy/pypy/src/578667b3fef9/rpython/translator/c/src/thread_pthread.c#lines-443:465
  https://bitbucket.org/pypy/pypy/src/5b42890d48c3/rpython/translator/c/src/thread_pthread.c#lines-443:465

Signalling condition outside of corresponding mutex is considered OK by
POSIX, but in Python context it can lead to at least memory corruption if we
consider the whole lifetime of python level lock. For example the following
logical scenario:

      T1                                          T2

  sema = Lock()
  sema.acquire()

                                              sema.release()

  sema.acquire()
  free(sema)

  ...

can translate to the next C-level calls:

      T1                                          T2

  # sema = Lock()
  sema = malloc(...)
  sema.locked = 0
  pthread_mutex_init(&sema.mut)
  pthread_cond_init (&sema.lock_released)

  # sema.acquire()
  pthread_mutex_lock(&sema.mut)
  # sees sema.locked == 0
  sema.locked = 1
  pthread_mutex_unlock(&sema.mut)

                                              # sema.release()
                                              pthread_mutex_lock(&sema.mut)
                                              sema.locked = 0
                                              pthread_mutex_unlock(&sema.mut)

                      # OS scheduler gets in and relinquishes control from T2
                      # to another process
                                              ...

  # second sema.acquire()
  pthread_mutex_lock(&sema.mut)
  # sees sema.locked == 0
  sema.locked = 1
  pthread_mutex_unlock(&sema.mut)

  # free(sema)
  pthread_mutex_destroy(&sema.mut)
  pthread_cond_destroy (&sema.lock_released)
  free(sema)

  # ...
  e.g. malloc() which returns memory where sema was

                                              ...
                      # OS scheduler returns control to T2
                      # sema.release() continues
                      #
                      # BUT sema was already freed and writing to anywhere
                      # inside sema block CORRUPTS MEMORY. In particular if
                      # _another_ python-level lock was allocated where sema
                      # block was, writing into the memory can have effect on
                      # further synchronization correctness and in particular
                      # lead to deadlock on lock that was next allocated.
                                              pthread_cond_signal(&sema.lock_released)

Note that T2.pthread_cond_signal(&sema.lock_released) CORRUPTS MEMORY as it
is called when sema memory was already freed and is potentially
reallocated for another object.

The fix is to move pthread_cond_signal to be done under corresponding mutex:

  # sema.release()
  pthread_mutex_lock(&sema.mut)
  sema.locked = 0
  pthread_cond_signal(&sema.lock_released)
  pthread_mutex_unlock(&sema.mut)

To do so this patch cherry-picks thread_pthread.h part of the following 3.2 commit:

commit 187aa545
Author: Kristján Valur Jónsson <kristjan@ccpgames.com>
Date:   Tue Jun 5 22:17:42 2012 +0000

    Signal condition variables with the mutex held.  Destroy condition variables
    before their mutexes.

 Python/ceval_gil.h      |  9 +++++----
 Python/thread_pthread.h | 15 +++++++++------
 2 files changed, 14 insertions(+), 10 deletions(-)

(ceval_gil.h is Python3 specific and does not apply to Python2.7)

The bug was there since 1994 - since at least [1]. It was discussed in 2001
with original code author[2], but the code was still considered to be
race-free. In 2010 the place where pthread_cond_signal should be - before or
after pthread_mutex_unlock - was discussed with the rationale to avoid
threads bouncing[3,4,5], and in 2012 pthread_cond_signal was moved to be
called from under mutex, but only for CPython3[6,7].

In 2019 the bug was (re-)discovered while testing Pygolang[8] on macOS with
CPython2 and PyPy2 and PyPy3.

[1] https://github.com/python/cpython/commit/2c8cb9f3d240
[2] https://bugs.python.org/issue433625
[3] https://bugs.python.org/issue8299#msg103224
[4] https://bugs.python.org/issue8410#msg103313
[5] https://bugs.python.org/issue8411#msg113301
[6] https://bugs.python.org/issue15038#msg163187
[7] https://github.com/python/cpython/commit/187aa545165d
[8] https://pypi.org/project/pygolang

(cherry picked from commit 187aa545)
Co-Authored-By: default avatarKristján Valur Jónsson <kristjan@ccpgames.com>
parent 403ca7ea
Fix race in PyThread_release_lock that was leading to memory corruption and
deadlocks. The fix applies to POSIX systems where Python locks are implemented
with mutex and condition variable because POSIX semaphores are either not
provided, or are known to be broken. One particular example of such system is
macOS.
......@@ -430,12 +430,15 @@ PyThread_free_lock(PyThread_type_lock lock)
(void) error; /* silence unused-but-set-variable warning */
dprintf(("PyThread_free_lock(%p) called\n", lock));
status = pthread_mutex_destroy( &thelock->mut );
CHECK_STATUS("pthread_mutex_destroy");
/* some pthread-like implementations tie the mutex to the cond
* and must have the cond destroyed first.
*/
status = pthread_cond_destroy( &thelock->lock_released );
CHECK_STATUS("pthread_cond_destroy");
status = pthread_mutex_destroy( &thelock->mut );
CHECK_STATUS("pthread_mutex_destroy");
free((void *)thelock);
}
......@@ -497,12 +500,12 @@ PyThread_release_lock(PyThread_type_lock lock)
thelock->locked = 0;
status = pthread_mutex_unlock( &thelock->mut );
CHECK_STATUS("pthread_mutex_unlock[3]");
/* wake up someone (anyone, if any) waiting on the lock */
status = pthread_cond_signal( &thelock->lock_released );
CHECK_STATUS("pthread_cond_signal");
status = pthread_mutex_unlock( &thelock->mut );
CHECK_STATUS("pthread_mutex_unlock[3]");
}
#endif /* USE_SEMAPHORES */
......
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