Why does Python threading.Condition() notify() require a lock?

Question

My question refers specifically to why it was designed that way, due to the unnecessary performance implication.

When thread T1 has this code:

cv.acquire()
cv.wait()
cv.release()

and thread T2 has this code:

cv.acquire()
cv.notify()  # requires that lock be held
cv.release()

what happens is that T1 waits and releases the lock, then T2 acquires it, notifies cv which wakes up T1. Now, there is a race-condition between T2's release and T1's reacquiring after returning from wait(). If T1 tries to reacquire first, it will be unnecessarily resuspended until T2's release() is completed.

Note: I'm intentionally not using the with statement, to better illustrate the race with explicit calls.

This seems like a design flaw. Is there any rationale known for this, or am I missing something?

Answer 1

This is not a definitive answer, but it's supposed to cover the relevant details I've managed to gather about this problem.

First, Python's threading implementation is based on Java's. Java's Condition.signal() documentation reads:

An implementation may (and typically does) require that the current thread hold the lock associated with this Condition when this method is called.

Now, the question was why enforce this behavior in Python in particular. But first I want to cover the pros and cons of each approach.

As to why some think it's often a better idea to hold the lock, I found two main arguments:

1. From the minute a waiter acquire()s the lock—that is, before releasing it on wait()—it is guaranteed to be notified of signals. If the corresponding release() happened prior to signalling, this would allow the sequence(where P=Producer and C=Consumer) P: release(); C: acquire(); P: notify(); C: wait() in which case the wait() corresponding to the acquire() of the same flow would miss the signal. There are cases where this doesn't matter (and could even be considered to be more accurate), but there are cases where that's undesirable. This is one argument.

2. When you notify() outside a lock, this may cause a scheduling priority inversion; that is, a low-priority thread might end up taking priority over a high-priority thread. Consider a work queue with one producer and two consumers (LC=Low-priority consumer and HC=High-priority consumer), where LC is currently executing a work item and HC is blocked in wait().

The following sequence may occur:

P                    LC                    HC
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
                     execute(item)                   (in wait())
lock()                                  
wq.push(item)
release()
                     acquire()
                     item = wq.pop()
                     release();
notify()
                                                     (wake-up)
                                                     while (wq.empty())
                                                       wait();

Whereas if the notify() happened before release(), LC wouldn't have been able to acquire() before HC had been woken-up. This is where the priority inversion occurred. This is the second argument.

The argument in favor of notifying outside of the lock is for high-performance threading, where a thread need not go back to sleep just to wake-up again the very next time-slice it gets—which was already explained how it might happen in my question.

Python's threading Module

In Python, as I said, you must hold the lock while notifying. The irony is that the internal implementation does not allow the underlying OS to avoid priority inversion, because it enforces a FIFO order on the waiters. Of course, the fact that the order of waiters is deterministic could come in handy, but the question remains why enforce such a thing when it could be argued that it would be more precise to differentiate between the lock and the condition variable, for that in some flows that require optimized concurrency and minimal blocking, acquire() should not by itself register a preceding waiting state, but only the wait() call itself.

Arguably, Python programmers would not care about performance to this extent anyway—although that still doesn't answer the question of why, when implementing a standard library, one should not allow several standard behaviors to be possible.

One thing which remains to be said is that the developers of the threading module might have specifically wanted a FIFO order for some reason, and found that this was somehow the best way of achieving it, and wanted to establish that as a Condition at the expense of the other (probably more prevalent) approaches. For this, they deserve the benefit of the doubt until they might account for it themselves.

Answer 2

There is no race condition, this is how condition variables work.

When wait() is called, then the underlying lock is released until a notification occurs. It is guaranteed that the caller of wait will reacquire the lock before the function returns (eg, after the wait completes).

You're right that there could be some inefficiency if T1 was directly woken up when notify() is called. However, condition variables are typically implemented via OS primitives, and the OS will often be smart enough to realize that T2 still has the lock, so it won't immediately wake up T1 but instead queue it to be woken.

Additionally, in python, this doesn't really matter anyways, as there's only a single thread due to the GIL, so the threads wouldn't be able to run concurrently anyways.

---

Additionally, it's preferred to use the following forms instead of calling acquire/release directly:

with cv:
    cv.wait()

And:

with cv:
    cv.notify()

This ensures that the underlying lock is released even if an exception occurs.

Answer 3

What happens is that T1 waits and releases the lock, then T2 acquires it, notifies cv which wakes up T1.

Not quite. The cv.notify() call does not wake the T1 thread: It only moves it to a different queue. Before the notify(), T1 was waiting for the condition to be true. After the notify(), T1 is waiting to acquire the lock. T2 does not release the lock, and T1 does not "wake up" until T2 explicitly calls cv.release().

Answer 4

A couple of months ago exactly the same question occurred to me. But since I had ipython opened, looking at threading.Condition.wait?? result (the source for the method) didn't take long to answer it myself.

In short, the wait method creates another lock called waiter, acquires it, appends it to a list and then, surprise, releases the lock on itself. After that it acquires the waiter once again, that is it starts to wait until someone releases the waiter. Then it acquires the lock on itself again and returns.

The notify method pops a waiter from the waiter list (waiter is a lock, as we remember) and releases it allowing the corresponding wait method to continue.

That is the trick is that the wait method is not holding the lock on the condition itself while waiting for the notify method to release the waiter.

UPD1: I seem to have misunderstood the question. Is it correct that you are bothered that T1 might try to reacquire the lock on itself before the T2 release it?

But is it possible in the context of python's GIL? Or you think that one can insert an IO call before releasing the condition, which would allow T1 to wake up and wait forever?

Answer 5

There are several reasons which are compelling (when taken together).

1. The notifier needs to take a lock

Pretend that Condition.notifyUnlocked() exists.

The standard producer/consumer arrangement requires taking locks on both sides:

def unlocked(qu,cv):  # qu is a thread-safe queue
  qu.push(make_stuff())
  cv.notifyUnlocked()
def consume(qu,cv):
  with cv:
    while True:       # vs. other consumers or spurious wakeups
      if qu: break
      cv.wait()
    x=qu.pop()
  use_stuff(x)

This fails because both the push() and the notifyUnlocked() can intervene between the if qu: and the wait().

Writing either of

def lockedNotify(qu,cv):
  qu.push(make_stuff())
  with cv: cv.notify()
def lockedPush(qu,cv):
  x=make_stuff()      # don't hold the lock here
  with cv: qu.push(x)
  cv.notifyUnlocked()

works (which is an interesting exercise to demonstrate). The second form has the advantage of removing the requirement that qu be thread-safe, but it costs no more locks to take it around the call to notify() as well.

It remains to explain the preference for doing so, especially given that (as you observed) CPython does wake up the notified thread to have it switch to waiting on the mutex (rather than simply moving it to that wait queue).

2. The condition variable itself needs a lock

The Condition has internal data that must be protected in case of concurrent waits/notifications. (Glancing at the CPython implementation, I see the possibility that two unsynchronized notify()s could erroneously target the same waiting thread, which could cause reduced throughput or even deadlock.) It could protect that data with a dedicated lock, of course; since we need a user-visible lock already, using that one avoids additional synchronization costs.

3. Multiple wake conditions can need the lock

(Adapted from a comment on the blog post linked below.)

def setSignal(box,cv):
  signal=False
  with cv:
    if not box.val:
      box.val=True
      signal=True
  if signal: cv.notifyUnlocked()
def waitFor(box,v,cv):
  v=bool(v)   # to use ==
  while True:
    with cv:
      if box.val==v: break
      cv.wait()

Suppose box.val is False and thread #1 is waiting in waitFor(box,True,cv). Thread #2 calls setSignal; when it releases cv, #1 is still blocked on the condition. Thread #3 then calls waitFor(box,False,cv), finds that box.val is True, and waits. Then #2 calls notify(), waking #3, which is still unsatisfied and blocks again. Now #1 and #3 are both waiting, despite the fact that one of them must have its condition satisfied.

def setTrue(box,cv):
  with cv:
    if not box.val:
      box.val=True
      cv.notify()

Now that situation cannot arise: either #3 arrives before the update and never waits, or it arrives during or after the update and has not yet waited, guaranteeing that the notification goes to #1, which returns from waitFor.

4. The hardware might need a lock

With wait morphing and no GIL (in some alternate or future implementation of Python), the memory ordering (cf. Java's rules) imposed by the lock-release after notify() and the lock-acquire on return from wait() might be the only guarantee of the notifying thread's updates being visible to the waiting thread.

5. Real-time systems might need it

Immediately after the POSIX text you quoted we find:

however, if predictable scheduling behavior is required, then that mutex shall be locked by the thread calling pthread_cond_broadcast() or pthread_cond_signal().

One blog post contains further discussion of the rationale and history of this recommendation (as well as of some of the other issues here).

Answer 6

It's explained in Python 3 documentation: https://docs.python.org/3/library/threading.html#condition-objects.

Note: the notify() and notify_all() methods don’t release the lock; this means that the thread or threads awakened will not return from their wait() call immediately, but only when the thread that called notify() or notify_all() finally relinquishes ownership of the lock.