pathikrit pathikrit - 2 months ago 12
Scala Question

Scala Partial Function Application Semantics + locking with synchronized

Based on my previous question on locks based on value-equality rather than lock-equality, I came up with the following implementation:

/**
* An util that provides synchronization using value equality rather than referential equality
* It is guaranteed that if two objects are value-equal, their corresponding blocks are invoked mutually exclusively.
* But the converse may not be true i.e. if two objects are not value-equal, they may be invoked exclusively too
* Note: Typically, no need to create instances of this class. The default instance in the companion object can be safely reused
*
* @param size There is a 1/size probability that two invocations that could be invoked concurrently is not invoked concurrently
*
* Example usage:
* import EquivalenceLock.{defaultInstance => lock}
* def run(person: Person) = lock(person) { .... }
*/
class EquivalenceLock(val size: Int) {
private[this] val locks = IndexedSeq.fill(size)(new Object())
def apply[U](lock: Any)(f: => U) = locks(lock.hashCode().abs % size).synchronized(f)
}

object EquivalenceLock {
implicit val defaultInstance = new EquivalenceLock(1 << 10)
}


I wrote some tests to verify that my lock functions as expected:

import EquivalenceLock.{defaultInstance => lock}

import scala.concurrent.Future
import scala.concurrent.ExecutionContext.Implicits.global
import scala.collection.mutable

val journal = mutable.ArrayBuffer.empty[String]

def log(msg: String) = journal.synchronized {
println(msg)
journal += msg
}

def test(id: String, napTime: Int) = Future {
lock(id) {
log(s"Entering $id=$napTime")
Thread.sleep(napTime * 1000L)
log(s"Exiting $id=$napTime")
}
}

test("foo", 5)
test("foo", 2)

Thread.sleep(20 * 1000L)

val validAnswers = Set(
Seq("Entering foo=5", "Exiting foo=5", "Entering foo=2", "Exiting foo=2"),
Seq("Entering foo=2", "Exiting foo=2", "Entering foo=5", "Exiting foo=5")
)

println(s"Final state = $journal")
assert(validAnswers(journal))


The above tests works as expected (tested over millions of runs). But, when I change the following line:

def apply[U](lock: Any)(f: => U) = locks(lock.hashCode().abs % size).synchronized(f)


to this:

def apply[U](lock: Any) = locks(lock.hashCode().abs % size).synchronized _


the tests fail.

Expected:

Entering foo=5
Exiting foo=5
Entering foo=2
Exiting foo=2


OR

Entering foo=2
Exiting foo=2
Entering foo=5
Exiting foo=5


Actual:

Entering foo=5
Entering foo=2
Exiting foo=2
Exiting foo=5


The above two pieces of code should be the same and yet the tests (i.e. the
lock(id)
always enters concurrently for the same
id
) for the second flavor (the one with partial application) of code. Why?

Answer

By default function parameters are evaluated eagerly. So

def apply[U](lock: Any) = locks(lock.hashCode().abs % size).synchronized _

is equivalent to

def apply[U](lock: Any)(f: U) = locks(lock.hashCode().abs % size).synchronized(f)

in this case f is evaluated before the synchronized block.

Comments