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- Question
What will be the output of the program?
class MyThread extends Thread { MyThread() { System.out.print(" MyThread"); } public void run() { System.out.print(" bar"); } public void run(String s) { System.out.println(" baz"); } } public class TestThreads { public static void main (String [] args) { Thread t = new MyThread() { public void run() { System.out.println(" foo"); } }; t.start(); } }
Options- A. foo
- B. MyThread foo
- C. MyThread bar
- D. foo bar
- Correct Answer
- MyThread foo
ExplanationOption B is correct because in the first line of main we're constructing an instance of an anonymous inner class extending from MyThread. So the MyThread constructor runs and prints "MyThread". The next statement in main invokes start() on the new thread instance, which causes the overridden run() method (the run() method defined in the anonymous inner class) to be invoked, which prints "foo"
Threads problems
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- 1. What will be the output of the program?
class s1 extends Thread { public void run() { for(int i = 0; i < 3; i++) { System.out.println("A"); System.out.println("B"); } } } class Test120 extends Thread { public void run() { for(int i = 0; i < 3; i++) { System.out.println("C"); System.out.println("D"); } } public static void main(String args[]) { s1 t1 = new s1(); Test120 t2 = new Test120(); t1.start(); t2.start(); } }
Options- A. Compile time Error There is no start() method
- B. Will print in this order AB CD AB...
- C. Will print but not be able to predict the Order
- D. Will print in this order ABCD...ABCD... Discuss
Correct Answer: Will print but not be able to predict the Order
Explanation:
We cannot predict the order in which threads are going to run.- 2. What will be the output of the program?
public class ThreadTest extends Thread { public void run() { System.out.println("In run"); yield(); System.out.println("Leaving run"); } public static void main(String []argv) { (new ThreadTest()).start(); } }
Options- A. The code fails to compile in the main() method
- B. The code fails to compile in the run() method
- C. Only the text "In run" will be displayed
- D. The text "In run" followed by "Leaving run" will be displayed Discuss
Correct Answer: The text "In run" followed by "Leaving run" will be displayed
- 3. The static method Thread.currentThread() returns a reference to the currently executing Thread object. What is the result of this code?
class Test { public static void main(String [] args) { printAll(args); } public static void printAll(String[] lines) { for(int i = 0; i < lines.length; i++) { System.out.println(lines[i]); Thread.currentThread().sleep(1000); } } }
Options- A. Each String in the array lines will output, with a 1-second pause.
- B. Each String in the array lines will output, with no pause in between because this method is not executed in a Thread.
- C. Each String in the array lines will output, and there is no guarantee there will be a pause because currentThread() may not retrieve this thread.
- D. This code will not compile. Discuss
Correct Answer: This code will not compile.
Explanation:
D. The sleep() method must be enclosed in a try/catch block, or the method printAll() must declare it throws the InterruptedException.A is incorrect, but it would be correct if the InterruptedException was dealt with.
B is incorrect, but it would still be incorrect if the InterruptedException was dealt with because all Java code, including the main() method, runs in threads.
C is incorrect. The sleep() method is static, so even if it is called on an instance, it still always affects the currently executing thread.
- 4. What will be the output of the program?
class s implements Runnable { int x, y; public void run() { for(int i = 0; i < 1000; i++) synchronized(this) { x = 12; y = 12; } System.out.print(x + " " + y + " "); } public static void main(String args[]) { s run = new s(); Thread t1 = new Thread(run); Thread t2 = new Thread(run); t1.start(); t2.start(); } }
Options- A. DeadLock
- B. It print 12 12 12 12
- C. Compilation Error
- D. Cannot determine output. Discuss
Correct Answer: It print 12 12 12 12
Explanation:
The program will execute without any problems and print 12 12 12 12.- 5. What will be the output of the program?
class s1 implements Runnable { int x = 0, y = 0; int addX() {x++; return x;} int addY() {y++; return y;} public void run() { for(int i = 0; i < 10; i++) System.out.println(addX() + " " + addY()); } public static void main(String args[]) { s1 run1 = new s1(); s1 run2 = new s1(); Thread t1 = new Thread(run1); Thread t2 = new Thread(run2); t1.start(); t2.start(); } }
Options- A. Compile time Error: There is no start() method
- B. Will print in this order: 1 1 2 2 3 3 4 4 5 5...
- C. Will print but not exactly in an order (e.g: 1 1 2 2 1 1 3 3...)
- D. Will print in this order: 1 2 3 4 5 6... 1 2 3 4 5 6... Discuss
Correct Answer: Will print but not exactly in an order (e.g: 1 1 2 2 1 1 3 3...)
Explanation:
Both threads are operating on different sets of instance variables. If you modify the code of the run() method to print the thread name it will help to clarify the output:- 6. Which two statements are true?
- Deadlock will not occur if wait()/notify() is used
- A thread will resume execution as soon as its sleep duration expires.
- Synchronization can prevent two objects from being accessed by the same thread.
- The wait() method is overloaded to accept a duration.
- The notify() method is overloaded to accept a duration.
- Both wait() and notify() must be called from a synchronized context.
Options- A. 1 and 2
- B. 3 and 5
- C. 4 and 6
- D. 1 and 3 Discuss
Correct Answer: 4 and 6
Explanation:
Statements (4) and (6) are correct. (4) is correct because the wait() method is overloaded to accept a wait duration in milliseconds. If the thread has not been notified by the time the wait duration has elapsed, then the thread will move back to runnable even without having been notified.(6) is correct because wait()/notify()/notifyAll() must all be called from within a synchronized, context. A thread must own the lock on the object its invoking wait()/notify()/notifyAll() on.
(1) is incorrect because wait()/notify() will not prevent deadlock.
(2) is incorrect because a sleeping thread will return to runnable when it wakes up, but it might not necessarily resume execution right away. To resume executing, the newly awakened thread must still be moved from runnable to running by the scheduler.
(3) is incorrect because synchronization prevents two or more threads from accessing the same object.
(5) is incorrect because notify() is not overloaded to accept a duration.
- 7. The following block of code creates a Thread using a Runnable target:
Runnable target = new MyRunnable(); Thread myThread = new Thread(target);
Which of the following classes can be used to create the target, so that the preceding code compiles correctly?
Options- A. public class MyRunnable extends Runnable{public void run(){}}
- B. public class MyRunnable extends Object{public void run(){}}
- C. public class MyRunnable implements Runnable{public void run(){}}
- D. public class MyRunnable implements Runnable{void run(){}} Discuss
Correct Answer: public class MyRunnable implements Runnable{public void run(){}}
Explanation:
The class correctly implements the Runnable interface with a legal public void run() method.Option A is incorrect because interfaces are not extended; they are implemented.
Option B is incorrect because even though the class would compile and it has a valid public void run() method, it does not implement the Runnable interface, so the compiler would complain when creating a Thread with an instance of it.
Option D is incorrect because the run() method must be public.
- 8. Which statement is true?
Options- A. If only one thread is blocked in the wait method of an object, and another thread executes the modify on that same object, then the first thread immediately resumes execution.
- B. If a thread is blocked in the wait method of an object, and another thread executes the notify method on the same object, it is still possible that the first thread might never resume execution.
- C. If a thread is blocked in the wait method of an object, and another thread executes the notify method on the same object, then the first thread definitely resumes execution as a direct and sole consequence of the notify call.
- D. If two threads are blocked in the wait method of one object, and another thread executes the notify method on the same object, then the first thread that executed the wait call first definitely resumes execution as a direct and sole consequence of the notify call. Discuss
Correct Answer: If a thread is blocked in the wait method of an object, and another thread executes the notify method on the same object, it is still possible that the first thread might never resume execution.
Explanation:
Option B is correct - The notify method only wakes the thread. It does not guarantee that the thread will run.Option A is incorrect - just because another thread activates the modify method in A this does not mean that the thread will automatically resume execution
Option C is incorrect - This is incorrect because as said in Answer B notify only wakes the thread but further to this once it is awake it goes back into the stack and awaits execution therefore it is not a "direct and sole consequence of the notify call"
Option D is incorrect - The notify method wakes one waiting thread up. If there are more than one sleeping threads then the choice as to which thread to wake is made by the machine rather than you therefore you cannot guarantee that the notify'ed thread will be the first waiting thread.
- 9. Which statement is true?
Options- A. The notifyAll() method must be called from a synchronized context.
- B. To call wait(), an object must own the lock on the thread.
- C. The notify() method is defined in class java.lang.Thread.
- D. The notify() method causes a thread to immediately release its locks. Discuss
Correct Answer: The notifyAll() method must be called from a synchronized context.
Explanation:
Option A is correct because the notifyAll() method (along with wait() and notify()) must always be called from within a synchronized context.Option B is incorrect because to call wait(), the thread must own the lock on the object that wait() is being invoked on, not the other way around.
Option C is wrong because notify() is defined in java.lang.Object.
Option D is wrong because notify() will not cause a thread to release its locks. The thread can only release its locks by exiting the synchronized code.
- 10. Which statement is true?
Options- A. A static method cannot be synchronized.
- B. If a class has synchronized code, multiple threads can still access the nonsynchronized code.
- C. Variables can be protected from concurrent access problems by marking them with the synchronized keyword.
- D. When a thread sleeps, it releases its locks. Discuss
Correct Answer: If a class has synchronized code, multiple threads can still access the nonsynchronized code.
Explanation:
B is correct because multiple threads are allowed to enter nonsynchronized code, even within a class that has some synchronized methods.A is incorrect because static methods can be synchronized; they synchronize on the lock on the instance of class java.lang.Class that represents the class type.
C is incorrect because only methods?not variables?can be marked synchronized.
D is incorrect because a sleeping thread still maintains its locks.
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- 1. What will be the output of the program?