Scala 部分函数应用语义 + 同步锁定

Question

基于基于值相等而不是锁相等的锁，我想出了以下实现：

/**
  * 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)
}

我写了一些测试来验证我的锁是否按预期运行：

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))

以上测试按预期工作（经过数百万次运行测试）。但是，当我更改以下行时：

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

对此：

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

测试失败。

预计：

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

或

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

实际：

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

上面的两段代码应该是相同的，但是对于第二种风格（部分应用的那个）的测试（即 lock(id) 总是同时进入相同的 id）的代码。为什么？

Answer 1

默认情况下，函数参数是急切求值的。所以

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

相当于

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

在这种情况下，f 在同步块之前计算。

Scala 部分函数应用语义 + 同步锁定

Scala Partial Function Application Semantics + locking with synchronized

scala

locking

synchronized

currying

partialfunction