具有start（）方法的Java多线程优先级关系

Question

My code and output are as below :-

My question is that the priority of main thread and thread t4 is 9 and that of thread t is 5 then why line 1 to 4 (marked in output) is coming before line 5 ie why t.start() gets priority over t4.start() which will be executed by main thread which has priority of 9.

Main thread priority is 9 so first t4.start() should get executed but then why t.start() is getting executed first ?

And if I reverse the order of calling start method ie if I call in order t4.start() and then t.start() then output is as expected.

Output

5

5

Main Thread Priority :9

5

Calling MyRunnable2 from main....9

Child Class                                - line 1

Child Class                                - line 2

End of Child loop                                - line 3

Calling MyRunnable2 from MyThread                - line 4

MyRunnable2 Class....Main....9                   - line 5

MyRunnable2 Class....From MyThread run....5

CODE :-

public class ThreadPriority 
{

    public static void main(String[] args) 
    {
        MyThread t = new MyThread();
        System.out.println(Thread.currentThread().getPriority());
        System.out.println(t.getPriority());
        t.setName("From MyThread");

        Thread.currentThread().setPriority(9);

        System.out.println("Main Thread Priority :" + Thread.currentThread().getPriority());
        System.out.println(t.getPriority());



        MyRunnable2 r4 = new MyRunnable2();
        Thread t4 = new Thread(r4,"Main");
        System.out.println("Calling MyRunnable2 from main"+"...."+t4.getPriority());

        t.start();
        t4.start();
    }

}


class MyRunnable2 implements Runnable
{
    public void run()
    {
        System.out.println("MyRunnable2 Class"+"...."+Thread.currentThread().getName()+"...."+Thread.currentThread().getPriority());

    }
}
class MyThread extends Thread
{
    public void run()
    {
        for (int i=0;i<2;i++)
            System.out.println("Child Class");

        System.out.println("End of Child loop");

        MyRunnable2 r2 = new MyRunnable2();
        Thread t2 = new Thread(r2,"From MyThread run");
        System.out.println("Calling MyRunnable2 from MyThread");
        t2.start();

    }

Answer 1

t4优先级更高，这并不意味着t4将首先完成，此外，jvm向OS询问线程，并将所有优先级设置在OS的较深层中，您只是设置了一个建议，它取决于很多事情！

Answer 2

The higher the integer, the higher the priority. At any given time, when multiple threads are ready to be executed, the runtime system chooses the runnable thread with the highest priority for execution. Only when that thread stops, yields, or becomes not runnable for some reason will a lower priority thread start executing.

If two threads of the same priority are waiting for the CPU, the scheduler chooses one of them to run in a round-robin fashion.

Rule of thumb : At any given time, the highest priority thread is running. However, this is not a guaranteed. The thread scheduler may choose to run a lower priority thread to avoid starvation. For this reason, use priority only to affect scheduling policy for efficiency purposes. Do not rely on thread priority for algorithm correctness.

I hope the above information clarifies your query.