(1)为什么是DE:Given:
1. public class Threads2 implements Runnable {
2.
3. public void nun() {
4. System.out.println(”run.”);
5. throw new RuntimeException(”Problem”);
6. }
7. public static void main(String[] args) {
8. Thread t = new Thread(new Threads2());
9. t.start();
10. System.out.println(”End of method.”);
11. }
12. }
Which two can be results? (Choose two.)
A. java.lang.RuntimeException: Problem
B. run.
java.lang.RuntimeException: Problem
C. End of method.
java.lang.RuntimeException: Problem
D. End of method.
run.
java.lang.RuntimeException: Problem
E. run.
java.lang.RuntimeException: Problem
End of method.
Answer: DE
(2)答案为什么是G?Click the Exhibit button.
Given:
1. public class TwoThreads {
2
3. private static Object resource = new Object();
4.
5. private static void delay(long n) {
6. try { Thread.sleep(n); }
7. catch (Exception e) { System.out.print(”Error “); }
8. }
9
10. public static void main(String[] args) {
11. System.out.print(”StartMain “);
12. new Thread1().start();
13. delay(1000);
14. Thread t2 = new Thread2();
15. t2.start();
16. delay(1000);
17. t2.interrupt
18. delay(1000);
19. System.out.print(”EndMain “);
20. }
21.
22. static class Thread 1 extends Thread {
23. public void run() {
24. synchronized (resource) {
25. System.out.print(”Startl “);
26. delay(6000);
27. System.out.print(”End1 “);
28. }
29. }
30. }
31.
32. static class Thread2 extends Thread {
33. public void run() {
34. synchronized (resource) {
35. System.out.print(”Start2 “);
36. delay(2000);
37. System.out.print(”End2 “);
38. }
39. }
40. }
41. }
Assume that sleep(n) executes in exactly m milliseconds, and all other
code executes in an insignificant amount of time. What is the output if
the main() method is run?
A. Compilation fails.
B. Deadlock occurs.
C. StartMain Start1 Error EndMain End1
D. StartMain Start1 EndMain End1 Start2 End2
E. StartMain Start1 Error Start2 EndMain End2 End1
F. StartMain Start1 Start2 Error End2 EndMain End1
G. StartMain Start1 EndMain End1 Start2 Error End2
Answer: G
1. public class Threads2 implements Runnable {
2.
3. public void nun() {
4. System.out.println(”run.”);
5. throw new RuntimeException(”Problem”);
6. }
7. public static void main(String[] args) {
8. Thread t = new Thread(new Threads2());
9. t.start();
10. System.out.println(”End of method.”);
11. }
12. }
Which two can be results? (Choose two.)
A. java.lang.RuntimeException: Problem
B. run.
java.lang.RuntimeException: Problem
C. End of method.
java.lang.RuntimeException: Problem
D. End of method.
run.
java.lang.RuntimeException: Problem
E. run.
java.lang.RuntimeException: Problem
End of method.
Answer: DE
(2)答案为什么是G?Click the Exhibit button.
Given:
1. public class TwoThreads {
2
3. private static Object resource = new Object();
4.
5. private static void delay(long n) {
6. try { Thread.sleep(n); }
7. catch (Exception e) { System.out.print(”Error “); }
8. }
9
10. public static void main(String[] args) {
11. System.out.print(”StartMain “);
12. new Thread1().start();
13. delay(1000);
14. Thread t2 = new Thread2();
15. t2.start();
16. delay(1000);
17. t2.interrupt
18. delay(1000);
19. System.out.print(”EndMain “);
20. }
21.
22. static class Thread 1 extends Thread {
23. public void run() {
24. synchronized (resource) {
25. System.out.print(”Startl “);
26. delay(6000);
27. System.out.print(”End1 “);
28. }
29. }
30. }
31.
32. static class Thread2 extends Thread {
33. public void run() {
34. synchronized (resource) {
35. System.out.print(”Start2 “);
36. delay(2000);
37. System.out.print(”End2 “);
38. }
39. }
40. }
41. }
Assume that sleep(n) executes in exactly m milliseconds, and all other
code executes in an insignificant amount of time. What is the output if
the main() method is run?
A. Compilation fails.
B. Deadlock occurs.
C. StartMain Start1 Error EndMain End1
D. StartMain Start1 EndMain End1 Start2 End2
E. StartMain Start1 Error Start2 EndMain End2 End1
F. StartMain Start1 Start2 Error End2 EndMain End1
G. StartMain Start1 EndMain End1 Start2 Error End2
Answer: G
main方法本身也是一个线程,当同时存在多个线程的时候,它们会竞争cpu,所以它们的执行顺序是没有保障的,就是说,哪个线程得到了cpu,哪个就可以执行,但是也不一定一直到线程结束,因为在执行的过程就有可能其他的线程夺去了.
有点头绪了
我总觉得main方法要等其他方法和线程执行完了,它排在最后。
这样认为就不对了第二题主要是
synchronized (resource)
这里锁定了,当Thread1先执行的时候,Thread2就只能等Thread1执行完后才能执行,因为他们锁定在同一对象上,反之亦然不理解的话google,baidu搜搜
try {
Thread.sleep(n);
} catch (Exception e) {
System.out.print("Error ");
}
} public static void main(String[] args) {
System.out.print("StartMain ");
new Thread1().start();
delay(1000);
Thread t2 = new Thread2();
t2.start();
delay(1000);
t2.interrupt();
delay(1000);
System.out.print("EndMain ");
} static class Thread1 extends Thread { public void run() {
synchronized (resource) {
System.out.print("Startl ");
delay(6000);
System.out.print("End1 ");
}
} } static class Thread2 extends Thread { public void run() {
synchronized (resource) {
System.out.print("Start2 ");
delay(2000);
System.out.print("End2 ");
}
}
}}
谢谢sagezk家里在装修没网,每次都是跑网吧来的
没软件
谢谢!