Lab Solution 5: rwlocks and Event Barriers
The lab checkoff sheet for all students can be found right here.
Get Started
Before starting, go ahead and clone the lab5
folder, which contains the test framework for the EventBarrier
class discussed in Problem 2.
git clone /usr/class/cs110/repos/lab5/shared lab5
Problem 1: Read-Write Locks
The following was a final exam question from a few years ago. The students were expected to write the code that follows.
The read-write lock (implemented by the rwlock
class) is a mutex
-like class with three public
methods:
class rwlock {
public:
rwlock();
void acquireAsReader();
void acquireAsWriter();
void release();
private:
// object state omitted
};
Any number of threads can acquire the lock as a reader without blocking one another. However, if a thread acquires the lock as a writer
, then all other acquireAsReader
and acquireAsWriter
requests block until the writer releases the lock. Waiting for the write lock will block until all readers release the lock so that the writer is guaranteed exclusive access to the resource being protected. This is useful if, say, you use some shared data structure that only very periodically needs to be modified. All reads from the data structure require you to hold the reader lock (so as many threads as you want can read the data structure at once), but any writes require you to hold the writer lock (giving the writing thread exclusive access).
The implementation ensures that as soon as one thread tries to get the writer lock, all other threads trying to acquire the lock—either as a reader or a writer—block until that writer gets the locks and releases it. That means the state of the lock can be one of three things:
- Ready, meaning that no one is trying to get the write lock.
- Pending, meaning that someone is trying to get the write lock but is waiting for all the readers to finish.
- Writing, meaning that someone is writing.
The following solution (written by Jerry Cain) relies on two mutex
es and two condition_variable_any
s. Here is the full interface for the rwlock
class:
class rwlock {
public:
rwlock(): numReaders(0), writeState(Ready) {}
void acquireAsReader();
void acquireAsWriter();
void release();
private:
int numReaders;
enum { Ready, Pending, Writing } writeState;
mutex readLock, stateLock;
condition_variable_any readCond, stateCond;
};
And here are the implementations of the three public
methods:
void rwlock::acquireAsReader() {
lock_guard<mutex> lgs(stateLock);
stateCond.wait(stateLock, [this]{ return writeState == Ready; });
lock_guard<mutex> lgr(readLock);
numReaders++;
}
void rwlock::acquireAsWriter() {
stateLock.lock();
stateCond.wait(stateLock, [this]{ return writeState == Ready; });
writeState = Pending;
stateLock.unlock();
lock_guard<mutex> lgr(readLock);
readCond.wait(readLock, [this]{ return numReaders == 0; });
writeState = Writing;
}
void rwlock::release() {
stateLock.lock();
if (writeState == Writing) {
writeState = Ready;
stateLock.unlock();
stateCond.notify_all();
return;
}
stateLock.unlock();
lock_guard<mutex> lgr(readLock);
numReaders--;
if (numReaders == 0) readCond.notify_one();
}
Very carefully study the implementation of the three methods, and answer the questions that appear below. This lab problem is designed to force you to really internalize the condition_variable_any
—one of the more difficult concepts in the entire threading segment of the course, and understand how it works.
- The implementation of
acquireAsReader
acquires thestateLock
(via thelock_guard
) before it does anything else, and it doesn’t release thestateLock
until the method exits. Why can’t the implementation be this instead?
void rwlock::acquireAsReader() {
stateLock.lock();
stateCond.wait(stateLock, [this]{ return writeState == Ready; });
stateLock.unlock();
lock_guard<mutex> lgr(readLock);
numReaders++;
}
-
The implementation of
acquireAsWriter
acquires thestateLock
before it does anything else and it releases thestateLock
just before it acquires thereadLock
. Why can’tacquireAsWriter
adopt the same approach asacquireAsReader
and just hold ontostateLock
until the method returns? -
Notice that we have a single
release
method instead ofreleaseAsReader
andreleaseAsWriter
methods. How does the implementation know if the thread acquired therwlock
as a writer instead of a reader (assuming proper use of the class)? -
The implementation of
release
relies onnotify_all
in one place andnotify_one
in another. Why are those the correct versions ofnotify
to call in each case? -
A thread that owns the lock as a reader might want to upgrade its ownership of the lock to that of a writer without releasing the lock first. Besides the fact that it’s a waste of time, what’s the advantage of not releasing the read lock before re-acquiring it as a writer, and how could the implementation of
acquireAsWriter
be updated so it can be called afteracquireAsReader
without an intervening release call?
Solution
- Why can't the implementation be the suggested alternative? Assume just two threads:
- Thread 1 calls
acquireAsReader
and is swapped off after third of five lines. - Thread 2 calls and progresses through all of
acquireAsWriter
. - Thread 1 progresses through rest of
acquireAsReader
. - We have one reader and one writer, and that’s forbidden.
- Thread 1 calls
- Why can't
acquireAsWriter
adopt the same approach asacquireAsReader
? If the writer doesn’t releasestateLock
before waiting for the number of readers to fall to 0, it blocks readers trying to release their locks from decrementingnumReaders
. - How does the implementation know if the thread acquired the
rwlock
as a writer instead of a reader? The implementation is such that the write state can only beWriting
when there’s one write lock and zero read locks. When the write state isWriting
, then only one thread could possibly be callingrelease
, unless the class is being used improperly. notify_one
vs.notify_all
answer: Any number of threads might be waiting for aReady
state so they can advance to acquire read locks. The writer needs to notify all of them when it releases the write lock. At most one thread—a writer—can be waiting for the number of readers to be 0, so callingnotify_one
is sufficient.- Not releasing the read lock before re-acquiring it as a writer, and updating the implementation:
- One advantage: fewer
mutex
es andcondition_variable_any
s need to be waited on, so the chance that the threads trying to upgrade the lock are forced to yield the processor is much, much smaller. - Another advantage: using the underlying thread (which are
pthread
s) and its support for priorities, you can give threads that are trying to upgrade higher priority so they get the processor before lower priority threads do. - Implementation idea: change the
acquireAsWriter
to accept abool
to state whether it holds a read lock already. - Better implementation idea: update the
rwlock
to maintain a set of thread ids (which are all the underlyingpthread_t
’s really are) that hold a read lock, and if the thread trying to upgrade finds its thread id in the set, then it knows it’s upgrading. It can then wait untilnumReaders == numUpgraders
instead ofnumReaders == 0
. Couple this with higher thread priorities and you can ensure that exactly one upgrading thread succeeds while all others wait. It’s true that all of the other upgraders technically have a read lock, but they’re blocked insideacquireAsWriter
, so they’re not actually reading whatever data structure is being accessed, and that's okay.
- One advantage: fewer
Problem 2: Event Barriers
An event barrier allows a group of one or more threads—we call them consumers—to efficiently wait
until an event occurs (i.e. the barrier is lift
ed by another thread, called the producer). The barrier is eventually restored by the producer, but only after consumers have detected the event, executed what they could only execute because the barrier was lifted, and notified the producer they’ve done what they need to do and moved past
the barrier. In fact, consumers and producers efficiently block (in lift
and past
, respectively) until all consumers have moved past the barrier. We say an event is in progress while consumers are responding to and moving past it.
The EventBarrier
implements this idea via a constructor and three zero-argument methods called wait
, lift
, and past
. The EventBarrier
requires no external synchronization, and maintains enough internal state to track the number of waiting consumers and whether an event is in progress. If a consumer arrives at the barrier while an event is in progress, wait
returns immediately without blocking.
The following test program (where all oslock
s and osunlock
s have been removed, for brevity) and sample run illustrate how the EventBarrier
works. The backstory for the sample run: singing minstrels approach a castle only to be blocked by a drawbridge gate. The minstrels wait until the gatekeeper lifts the gate, allowing the minstrels to cross. The gatekeeper only lowers the gate after all minstrels have crossed the bridge, and the minstrels only proceed toward the castle once all have crossed the bridge.
static void minstrel(const string& name, EventBarrier& eb) {
cout << oslock << name << " walks toward the drawbridge." << endl << osunlock;
sleep(random() % 3 + 3); // minstrels arrive at drawbridge at different times
cout << oslock << name << " arrives at the drawbridge, must wait." << endl << osunlock;
eb.wait(); // all minstrels wait until drawbridge has been raised
cout << oslock << name << " detects drawbridge lifted, starts crossing." << endl << osunlock;
sleep(random() % 3 + 2); // minstrels walk at different rates
cout << oslock << name << " has crossed the bridge." << endl << osunlock;
eb.past();
cout << oslock << name << " carries on, knowing all others have crossed." << endl << osunlock;
}
static void gatekeeper(EventBarrier& eb) {
sleep(random() % 5 + 7);
cout << oslock << "Gatekeeper raises the drawbridge." << endl << osunlock;
eb.lift(); // lift the drawbridge
cout << oslock << "Gatekeeper lowers drawbridge knowing all have crossed." << endl << osunlock;
}
static string kMinstrelNames[] = {
"Peter", "Paul", "Mary", "Manny", "Moe", "Jack"
};
static const size_t kNumMinstrels = 6;
int main(int argc, char *argv[]) {
EventBarrier drawbridge;
thread minstrels[kNumMinstrels];
for (size_t i = 0; i < kNumMinstrels; i++) {
minstrels[i] = thread(minstrel, kMinstrelNames[i], ref(drawbridge));
}
thread g(gatekeeper, ref(drawbridge));
for (thread& c: minstrels) c.join();
g.join();
return 0;
}
myth52$ ./event-barrier-test
Peter walks toward the drawbridge.
Paul walks toward the drawbridge.
Manny walks toward the drawbridge.
Moe walks toward the drawbridge.
Mary walks toward the drawbridge.
Jack walks toward the drawbridge.
Peter arrives at the drawbridge, must wait.
Paul arrives at the drawbridge, must wait.
Manny arrives at the drawbridge, must wait.
Mary arrives at the drawbridge, must wait.
Jack arrives at the drawbridge, must wait.
Moe arrives at the drawbridge, must wait.
Gatekeeper raises the drawbridge.
Peter detects drawbridge lifted, starts crossing.
Paul detects drawbridge lifted, starts crossing.
Manny detects drawbridge lifted, starts crossing.
Mary detects drawbridge lifted, starts crossing.
Jack detects drawbridge lifted, starts crossing.
Moe detects drawbridge lifted, starts crossing.
Peter has crossed the bridge.
Paul has crossed the bridge.
Manny has crossed the bridge.
Jack has crossed the bridge.
Mary has crossed the bridge.
Moe has crossed the bridge.
Moe carries on, knowing all others have crossed.
Gatekeeper lowers drawbridge knowing all have crossed.
Peter carries on, knowing all others have crossed.
Paul carries on, knowing all others have crossed.
Manny carries on, knowing all others have crossed.
Jack carries on, knowing all others have crossed.
Mary carries on, knowing all others have crossed.
Your lab5
folder includes event-barrier.h
, event-barrier.cc
, and ebtest.cc
, and typing make
should generate an executable called ebtest
that you can run to ensure that the EventBarrier
class you'll flesh out in event-barrier.h
and .cc
are working properly. The one exercise in this lab that has you do any coding is this one, as it expects you complete the implementation stub you've been supplied with.
Solution
// .h
class EventBarrier {
public:
EventBarrier();
void wait();
void lift();
void past();
private:
size_t numWaiting;
bool occurring;
std::mutex m;
std::condition_variable_any cv;
};
// .cc
EventBarrier::EventBarrier(): numWaiting(0), occurring(false) {}
void EventBarrier::wait() {
lock_guard<mutex> lg(m);
numWaiting++;
cv.wait(m, [this] { return occurring; });
}
void EventBarrier::lift() {
lock_guard<mutex> lg(m);
occurring = true;
cv.notify_all();
cv.wait(m, [this] { return numWaiting == 0; });
occurring = false;
}
void EventBarrier::past() {
lock_guard<mutex> lg(m);
numWaiting--;
if (numWaiting > 0) {
cv.wait(m, [this] { return numWaiting == 0; });
} else {
cv.notify_all();
}
}
Checkoff Questions Solutions
-
Problem 1a: The implementation of
acquireAsReader
acquires thestateLock
(via thelock_guard
) before it does anything else, and it doesn't release thestateLock
until the method exits. The lab handout presents an alternate implementation ofacquireAsReader
where thestateLock
is released even sooner. Why doesn't that work?- Thread 1 calls
acquireAsReader
and is swapped off after third of five lines. - Thread 2 calls and progresses through all of
acquireAsWriter
. - Thread 1 progresses through rest of
acquireAsReader
. - We have one reader and one writer, and that’s forbidden.
- Thread 1 calls
-
Problem 1b: The implementation of
rwlock::release
relies onnotify_all
in one place andnotify_one
in another. Why are those the correct versions ofnotify
to call in each case? Any number of threads might be waiting for aReady
state so they can advance to acquire read locks. The writer needs to notify all of them when it releases the write lock. At most one thread—a writer—can be waiting for the number of readers to be 0, so callingnotify_one
is sufficient. -
Problem 2a: An
EventBarrier
causes threads to block in various places - inwait
, if the block has not yet been lifted, inlift
, if not all waiting have passed yet, and inpast
, if all others waiting have not yet passed. How are each of these blocks implemented in code? Each of these is implemented via a condition variable. Forwait
, untiloccurring
is true, we wait to be signaled by the barrier being lifted. Forlift
, untilnumWaiting
is 0, we wait to be signaled by a thread going past the barrier. And forpast
, untilnumWaiting
is 0, we wait to be signaled by the last thread that goes past the barrier. -
Problem 2b: There are cases where
wait
andlift
never block. For example,wait
will not block if the event is already occurring.lift
will never block if there are none waiting for the event to happen. How is it possible for these to not block, given that we always callcv.wait
in those places?cv.wait
with a lambda function calls the actualcv.wait(mutex)
in awhile
loop with that lambda as the condition to break. So if the condition is true when we initially callcv.wait
, we will never actually wait, since we will never loop.
Icons by Piotr Kwiatkowski