Implicit synchronization in java?

Question

when i have a method like

public void unsynchronizedMethod(){
  callSynchronizedMethod();
  //dosomestuff after this line
}

Does this mean that all content after calling the SynchronizedMethod () method in the unsynchronizedMethod () method is implicitly synchronized?

Answer 1

Synchronization is determined in a simple way:

• For the method synchronized: when executing this method
• For a block synchronized: when executing this block

Here are the relevant excerpts from the Java Language Specification 3rd Edition:

JLS 14.19 Statement synchronized

A synchronized:

• gets a mutual exclusion lock on behalf of an executable thread,
• executes a block
• then releases the lock.

, .

synchronized synchronized, (§JLS 8.4.3.6. this ( ) Class , ( a static, this static).

, , :

public void unsynchronizedMethod(){
  callSynchronizedMethod();

  doSomeUnsynchronizedStuff(); // no "implicit synchronization" here
}

, : . .

.

• Java 2nd Edition, 67:

• Java?
• ?

Answer 2

. callSynchronizedMethod().

Answer 3

, .

, .

callSynchronizedMethod() , .

Answer 4

, . . , , .

. 0.

class example extends Thread{
   //Global value updated by the example threads
   public static volatile int value= 0;
    public void run(){
    while(true)//run forever
       unsynchMethod();

   }
   public void unsynchMethod(){
     synchronizedMethod();
    //Count value up and then back down to 0
    for(int i =0; i < 20000;++i)
       value++;//reads increments and updates value (3 steps)
    for(int i = 0; i < 20000;++i)
       value--;//reads decrements and updates value (3 steps)
      //Print value 
      System.out.println(value);
   }
   //Classlevel synchronized function
   public static synchronized void synchronizedMethod(){
      //not important
   }
   public static void main(String... args){
    example a = new example();
    example b = new example();
    a.start();
    b.start();
   }

}

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