CODEWITHSUNDEEP
multithreadingscalasynchronized
user2547081
user2547081

Reputation: 189

difficulty in understanding the synchronized definition

I am learning to write code in multi-threading. I understand the literal meaning of synchronized operator in Scala. But, when I run the following code, I do not understand the output.

package examples
import scala.concurrent.ops._

object concurrent extends App {

    class BoundedBuffer[A](N:Int) {
        var in = 0;
        var out = 0;
        var n = 0;

        def put(x:A) = synchronized {
          while(n>=N)
            wait()
          in = (in + 1)/N;
          n = n+1;
          println("In put.")
          if(n==1)
            notifyAll()
        }

        def get = synchronized {
          while(n==0)
            wait()
          out = (out + 1)%N;
          n = n-1;
          println("In get.")
          if(n == N-1)
            notifyAll()
        }
    }

    val buf = new BoundedBuffer[Int](10)
    spawn {
        while(true)
            buf.put(0);
    }
    spawn {
        while(true)
            buf.get;
    }
}

with synchronized in put and get, the function would continue forever, which is expected. But, when I remove the synchronized from the definition, the output will be 

In put.
In put.
In get.
In get.

Could Anyone explain why the results look like this? Thanks a lot.

Upvotes: 0

Views: 94

Answers (1)

DaoWen
DaoWen

Reputation: 33029

The JVM's memory model has no guarantees for sequential consistency if you don't use something like volatile or synchronized. This means that each thread essentially has an independent view of the current value of n.

My guess is that something like this is happening:

  1. The first thread spawns
  2. First thread runs put twice and gets stuck waiting (I guess N=2?)
  3. The second thread spawns, getting the current view of n=2
  4. Second thread runs get twice, bringing n down to 0, and gets stuck waiting
  5. Since there's no synchronization construct here each thread maintains its private view of the value of n, and it never changes—they just stay stuck in the while loops

Try making n volatile and see what happens. (I don't that will give you 100% correct behavior, but I don't think it will get stuck either.)

Upvotes: 1

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