About the start call problem

package com.test;
  
  public class MyThread extends Thread
  {
  	
  	@Override
  	public void run() {
  		System.out.println("run函数");
  	}
  	
  	public void disp1()
  	{
  		System.out.println("disp函数1");
  	}
  	
  	public void disp2()
  	{
  		System.out.println("disp函数2");
  	}
  }

package com.test;
  public class TestBlockingQueue {
  		
  	public static void main(String[] args) 
      {
  		MyThread mt = new MyThread();	
  		mt.start();
  		mt.disp1();				
      }
  }

write this , why that run () function in disp1 () after execution ? Principle is what ?
------ Solution ---------------------------------------- ----
changed following a look at the results.

  @Override
      public void run() {
  for (int i = 0; i < 10000; i++) {
          System.out.println("run函数");
  }
      }
       
      public void disp1()
      {
  for (int i = 0; i < 10000; i++) {
          System.out.println("disp函数1");
  }
      }
  

------ Solution ------------------------------------- -------
thread's call , the order of execution is managed with a JVM .
------ Solution -------------------------------------- ------

  mt.start(); // 当主线程可以理解为main，执行到这里，子线程mt起来之后，mt执行，main继续执行。这两个是同时运行的
          mt.disp1();//到这里的话打印哪个都是不确定的。哪个快就执行哪个打印。
  

procedures according to the first floor you can see out.

------ Solution ------------------------------------ --------
Well, this order does not determine .
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random ? Or a regular ?
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me yesterday that no circular pen questions , you would like to explain this matter there is no fixed order of the thread's call it ?
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oh , is uncertain , right !
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OK