大侠们帮我看看多线程方法同步问题

Question

如下java代码：package com.test;public class ThreadTest implements Runnable { private static int i = 0;
private static int j = 0; public void run() {
f1();
f2();
} private void f1() {
synchronized (ThreadTest.class) {
try {
Thread.sleep(10);
} catch (InterruptedException e) {
e.printStackTrace();
}
System.out.println("currentThread:"
+ Thread.currentThread().getId());
System.out.println("i:" + (++i) + "currentThread:"
+ Thread.currentThread().getId());
} } private void f2() {
synchronized (ThreadTest .class) {
try {
Thread.sleep(10);
} catch (InterruptedException e) {
e.printStackTrace();
}
System.out.println("\tcurrentThread:"
+ Thread.currentThread().getId());
System.out.println("j:" + (++j) + "\tcurrentThread:"
+ Thread.currentThread().getId());
} } public static void main(String[] args) { Thread th1 = new Thread(new ThreadTest ());
Thread th2 = new Thread(new ThreadTest ());
Thread th3 = new Thread(new ThreadTest ());
Thread th4 = new Thread(new ThreadTest ());
th1.start();
th2.start();
th3.start();
th4.start(); }
}
为什么f1()和f2()的第二行打印是同步的，而第一行却不同步？注释掉第一行，打印结果如下：
i:1 currentThread:8
j:1 currentThread:8
i:2 currentThread:11
j:2 currentThread:11
i:3 currentThread:9
j:3 currentThread:9
i:4 currentThread:10
j:4 currentThread:10注释掉第二行，打印结果如下：
currentThread:8
currentThread:11
currentThread:11
currentThread:9
currentThread:9
currentThread:10
currentThread:8
currentThread:10

Answer 1

f1()执行完之后会重新去竞争琐,所以f2()的执行顺序取决于虚拟机如果把f1和f2的锁去掉,把run()方法改为:    public void run() {
   synchronized(ThreadTest.class) {
           f1();
           f2();
          }
      }   就会f1()和f2()就会顺序执行了