Read many, write exclusive locks – RwLock

Consider a situation where you have a resource that must be manipulated only a single thread at a time, but is safe to be queried by many—that is, you have many readers and only one writer.

While you could protect this resource with a mutex , the trouble is that the mutex makes no distinction between its lockers; every thread will be forced to wait, no matter what their intentions. RwLock<T> is an alternative to the mutex concept, allowing for two kinds of locks—read and write. Analogously to Rust’s references, there can only be one write lock taken at a time but multiple reader locks, exclusive of a write lock. Here’s an example:

use std::thread; use std::sync::{Arc, RwLock}; fn main () { let resource: Arc < RwLock < u16 >> = Arc::new(RwLock::new( 0 )); let total_readers = 5 ; let mut reader_jhs = Vec::with_capacity(total_readers); for _ in 0 ..total_readers { let resource = Arc::clone( & resource); reader_jhs.push(thread::spawn( move || { let mut total_lock_success = 0 ; let mut total_lock_failure = 0 ; let mut total_zeros = 0 ; while total_zeros < 100 { match resource.try_read() { Ok(guard) => { total_lock_success += 1 ; if * guard == 0 { total_zeros += 1 ; } } Err(_) => { total_lock_failure += 1 ; } } } (total_lock_failure, total_lock_success) })); } { let mut loops = 0 ; while loops < 100 { let mut guard = resource.write().unwrap(); * guard = guard.wrapping_add( 1 ); if * guard == 0 { loops += 1 ; } } } for jh in reader_jhs { println ! ( "{:?}" , jh.join().unwrap()); } }

The idea here is that you’ll have one writer thread spinning and incrementing, in a wrapping fashion, a shared resource—a u16 . Once the u16 has been wrapped 100 times, the writer thread will exit. Meanwhile, a total_readers number of read threads will attempt to take a read lock on the shared resource—a u16—until it hits zero 100 times. You’re gambling here, essentially, on thread ordering. Quite often, the program will exit with this result:

( 0 , 100 ) ( 0 , 100 ) ( 0 , 100 ) ( 0 , 100 ) ( 0 , 100 )

This means that each reader thread never failed to get its read lock—there were no write locks present. This means that the reader threads were scheduled before the writer. Your main function only joins on reader handlers, so the writer is left writing as you exit. Sometimes, you’ll hit just the right scheduling order and get the following result:

( 0 , 100 ) ( 126143752 , 2630308 ) ( 0 , 100 ) ( 0 , 100 ) ( 126463166 , 2736405 )

In this particular instance, the second and final reader threads were scheduled just after the writer and managed to catch a time when the guard was not zero. Recall that the first element of the pair is the total number of times the reader thread was not able to get a read lock and was forced to retry.

The second is the number of times that the lock was acquired. In total, the writer thread did (2^18 * 100) ~= 2^24 writes, whereas the second reader thread did log_2 2630308 ~= 2^21 reads. That’s a lot of lost writes which maybe okay. Of more concern, that’s a lot of useless loops, approximately 2^26 .

How do you avoid all this wasted effort? Well, like most things, it depends on what you’re trying to do. If you need every reader to get every write, then an MPSC is a reasonable choice. It would look like this:

use std::thread; use std::sync::mpsc; fn main () { let total_readers = 5 ; let mut sends = Vec::with_capacity(total_readers); let mut reader_jhs = Vec::with_capacity(total_readers); for _ in 0 ..total_readers { let (snd, rcv) = mpsc::sync_channel( 64 ); sends.push(snd); reader_jhs.push(thread::spawn( move || { let mut total_zeros = 0 ; let mut seen_values = 0 ; for v in rcv { seen_values += 1 ; if v == 0 { total_zeros += 1 ; } if total_zeros >= 100 { break ; } } seen_values })); } { let mut loops = 0 ; let mut cur: u16 = 0 ; while loops < 100 { cur = cur.wrapping_add( 1 ); for snd in & sends { snd.send(cur).expect( "failed to send" ); } if cur == 0 { loops += 1 ; } } } for jh in reader_jhs { println ! ( "{:?}" , jh.join().unwrap()); } }

It will run for a while and print out the following:

6553600 6553600 6553600 6553600 6553600

But what if every reader does not need to see every write, meaning that it’s acceptable for a reader to miss writes so long as it does not miss all of the writes? You have options.

Blocking until conditions change – condvar

One such option is a condvar, or CONDition VARiable. Condvars are a nifty way to block a thread, pending a change in some Boolean condition. One difficulty is that condvars are associated exclusively with mutexes, but in this example, you need not mind it all that much.

The way a condvar works is that after taking a lock on a mutex, you pass the MutexGuard into Condvar::wait , which blocks the thread. Other threads may go through this same process, blocking on the same condition. Some other thread will take the same exclusive lock and eventually call either notify_one or notify_all on the condvar.

The first wakes up a single thread; the second wakes up all threads. Condvars are subject to spurious wakeup, meaning the thread may leave its block without a notification being sent to it. For this reason, condvars check their conditions in a loop. But, once the condvar wakes, you are guaranteed to hold the mutex, which prevents deadlocks on spurious wakeup.

Now adapt your example to use a condvar:

use std::thread; use std::sync::{Arc, Condvar, Mutex}; fn main () { let total_readers = 5 ; let mutcond: Arc < (Mutex < ( bool , u16 ) > , Condvar) > = Arc::new((Mutex::new(( false , 0 )), Condvar::new()));

You’re synchronizing threads on mutcond , which is an Arc<(Mutex<(bool, u16)>, Condvar)> . Rust’s condvar is touch awkward. It’s undefined behavior to associate a condvar with more than one mutex, but there’s really nothing in the type of Condvar that makes that an invariant. You just have to remember to keep them associated.

To that end, it’s not uncommon to see a Mutex and Condvar paired up in a tuple, as here. Now, why Mutex<(bool, u16)>? The second element of the tuple is your resource, which is common to other examples. The first element is a Boolean flag, which you use as a signal to mean that there are writes available. Here are your reader threads:

let mut reader_jhs = Vec::with_capacity(total_readers); for _ in 0 ..total_readers { let mutcond = Arc::clone( & mutcond); reader_jhs.push(thread::spawn( move || { let mut total_zeros = 0 ; let mut total_wakes = 0 ; let & ( ref mtx, ref cnd) = &* mutcond; while total_zeros < 100 { let mut guard = mtx.lock().unwrap(); while ! guard. 0 { guard = cnd.wait(guard).unwrap(); } guard. 0 = false ; total_wakes += 1 ; if guard. 1 == 0 { total_zeros += 1 ; } } total_wakes })); }

Until total_zeros hits 100, the reader thread locks the mutex, checks the guard inside the mutex for writes availability, and, if there are no writes, does a wait on the condvar, which gives up the lock. The reader thread is then blocked until a notify_all is called—as you’ll see shortly.

Every reader thread then races to be the first to reacquire the lock. The lucky winner notes that there are no more writes to be read and then does the normal flow. It bears repeating that every thread that wakes up from a condition wait is racing to be the first to reacquire the mutex.

Your reader is uncooperative in that; it immediately prevents the chance of any other reader threads finding a resource available. However, they will still wake up spuriously and be forced to wait again. Maybe! The reader threads are also competing with the writer thread to acquire the lock. Now, look at the writer thread:

let _ = thread::spawn( move || { let & ( ref mtx, ref cnd) = &* mutcond; loop { let mut guard = mtx.lock().unwrap(); guard. 1 = guard. 1. wrapping_add( 1 ); guard. 0 = true ; cnd.notify_all(); } });

The writer thread is an infinite loop, which you orphan in an unjoined thread. Now, it’s entirely possible that the writer thread will acquire the lock, bump the resource, notify waiting reader threads, give up the lock, and then immediately re-acquire the lock to begin the whole process before any reader threads can get scheduled in.

This means it’s entirely possible that the resource being zero will happen several times before a reader thread is lucky enough to notice. Now close this program:

for jh in reader_jhs { println ! ( "{:?}" , jh.join().unwrap()); } }

Ideally, what you’d like is some manner of bi-directionality—you’d like the writer to signal that there are reads and the reader to signal that there is capacity. You might, for instance, layer another condition variable into the mix, this one for the writer, but that’s not what you have here and the program suffers for it. Here’s one run:

7243473 6890156 6018468 6775609 6192116

Phew! That’s significantly more loops than the previous example. None of this is to say that condition variables are hard to use—they’re not—it’s just that they need to be used in conjunction with other primitives.