为什么两个线程产生以下结果（ocjp 测试）？

Question

I have below ocjp test which i don't understand the answer (as I thought the answer should be A+D,but the correct answer is A+B), anyone can explain that?:

Given a piece of code. Which two are possible results? (Select two)

 public class Cruiser { private int a = 0; public void foo() { Runnable r = new LittleCruiser(); new Thread(r).start(); new Thread(r).start(); } public static void main(String arg[]) { Cruiser c = new Cruiser(); c.foo(); } public class LittleCruiser implements Runnable { public void run() { int current = 0; for (int i = 0; i < 4; i++) { current = a; System.out.print(current + ", "); a = current + 2; } } } }

A) 0, 2, 4, 0, 2, 4, 6, 6,
B) 0, 2, 4, 6, 8, 10, 12, 14,
C) 0, 2, 4, 6, 8, 10, 2, 4,
D) 0, 0, 2, 2, 4, 4, 6, 6, 8, 8, 10, 10, 12, 12, 14, 14,
E) 0, 2, 4, 6, 8, 10, 12, 14, 0, 2, 4, 6, 8, 10, 12, 14,

Answers: A, B

Answer 1

The run method has a for loop that executes 4 times and prints something each time. As you run 2 threads, you should end up printing 8 things, so answers D and E are not possible.

As for what is being displayed, you need to see that a is shared and remember that the threads run in parallel: it's possible that run from thread 1 runs entirely and then the run of thread 2 runs entirely, or they get mixed up. In the latter case, thread 1 may be inside the for loop printing something, and then instead of executing the +2 you switch to thread 2 and print again the same number.

Answer A is possible: thread 1 starts and executes current = a; in the for loop, so current is equal to 0 and a hasn't been modified. Then thread 2 starts and executes the for 3 times, thus printing 0, 2, 4, . Let's say it also sets a to 6 . Then we're back to thread 1 that continues with its current being equal to 0 , and it has time to print 3 times, thus printing 0, 2, 4, . It also sets a to 6 . Then we're back on thread 2 that gets a and prints it ie '6,', then we're back on thread that also gets a` and prints it ie '6,'.

Answer B is possible: for ex thread 1 runs entirely then thread 2. Or the threads get mixed up but not too much, eg each thread has time to execute the 3 lines inside the for loop before the context switches to the other thread.

Answer C is not possible: you can only go back to a previous number if you do a current = a . As you go back to 2 , it means that thread 1 already printed 0 , then did the +2 , then current = a . Then thread 2 runs entirely and prints 2, 4, 6, 8 . Then thread 1 goes back to printing 2 . It's not possible to be able to print 10 then go back to 2 (you could go back to 4 though).

Answer 2

it's because private int a = 0; it's shared across the two threads. So both threads are interacting with this variable at the same time. The following is likely to happen:

0, 2, 4, 6, 8, 10, 12, 14

0, 0, 2, 4, 2, 6, 4, 6,

0, 2, 4, 6, 0, 2, 4, 6,