Why does the method start not called?

Question

I have 2 questions: 1. Why run () is not called when the program starts 2. If run () is called, will randomScore change the value?

    import java.*;


public class StudentThread extends Thread {
    int ID;  
    public static volatile int randomScore;   //will it change the value of randomScore defined here?
      StudentThread(int i) {
          ID = i;
      }

      public void run() {
          randomScore = (int)( Math.random()*1000);
      }

public static void main(String args[]) throws Exception {
    for (int i = 1;i< 10 ;i++) 
    {
            StudentThread student = new StudentThread(5);
            student.start(); 
            System.out.println(randomScore);
    }             
}  
 }

Answer 1

Most importantly, you need to change

randomScore = (int) Math.random() * 1000;

to

randomScore = (int) (Math.random() * 1000);
                    ^                    ^

since it (int) Math.random()will always be 0.

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Another important thing to note is that the main thread continues and prints the value randomScore, without waiting for another thread to change the value. Try adding Thread.sleep(100);after startor student.join()to wait until the student stream ends.

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You should also be aware that the Java memory model allows threads to cache their values ​​for variables. Perhaps the main thread has cached its own value.

randomScore volatile:

public static volatile int randomScore;

Answer 2

• , . student.join() student.start()
• run(), randomScore, , @aioobe .

Answer 3

, , . , .

, ​​ , CountdownLatch ( 1 , latch.await()) , , Thread.join(), .

Answer 4

, Math.random int 1000.

  randomScore = (int) (Math.random()*1000);

, :

• (inital state)

• • Math.random , 0 1
  • double int, 0
  • 0 1000, 0.

, .

TL; DR:

• run()

• , , . , 0 run.

Answer 5

Yes, the method is being called run. But it may be too fast. You can join threadwith the main one threadand wait a while.