concurrency - Volatile in java -

as far know volatile write happens-before volatile read, see freshest data in volatile variable. question concerns term happens-before , take place? wrote piece of code clarify question.

class test {    volatile int a;    public static void main(string ... args) {      final test t = new test();      new thread(new runnable(){         @override         public void run() {             thread.sleep(3000);             t.a = 10;         }      }).start();      new thread(new runnable(){         @override         public void run() {             system.out.println("value " + t.a);         }      }).start();    } } 

(try catch block omitted clarity)

in case see value 0 printed on console. without thread.sleep(3000); see value 10. case of happens-before relationship or prints 'value 10' because thread 1 starts bit earlier thread 2?

it great see example behaviour of code , without volatile variable differs in every program start, because result of code above depends only(at least in case) on order of threads , on thread sleeping.

you see value 0 because read executed before write. , see value 10 because write executed before read.

if want have test more unpredictable output, should have both of threads await countdownlatch, make them start concurrently:

final countdownlatch latch = new countdownlatch(1); new thread(new runnable(){     @override     public void run() {         try {             latch.await();             t.a = 10;         }         catch (interruptedexception e) {             // end thread         }     }  }).start();  new thread(new runnable(){     @override     public void run() {         try {             latch.await();             system.out.println("value " + t.a);         }         catch (interruptedexception e) {             // end thread         }     }  }).start();  thread.sleep(321); // go  latch.countdown();