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- Question
What will be the output of the program?
class Test116 { static final StringBuffer sb1 = new StringBuffer(); static final StringBuffer sb2 = new StringBuffer(); public static void main(String args[]) { new Thread() { public void run() { synchronized(sb1) { sb1.append("A"); sb2.append("B"); } } }.start(); new Thread() { public void run() { synchronized(sb1) { sb1.append("C"); sb2.append("D"); } } }.start(); /* Line 28 */ System.out.println (sb1 + " " + sb2); } }
Options- A. main() will finish before starting threads.
- B. main() will finish in the middle of one thread.
- C. main() will finish after one thread.
- D. Cannot be determined.
- Correct Answer
- Cannot be determined.
ExplanationCan you guarantee the order in which threads are going to run? No you can't. So how do you know what the output will be? The output cannot be determined.add this code after line 28:
try { Thread.sleep(5000); } catch(InterruptedException e) { }
and you have some chance of predicting the outcome.
Threads problems
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- 1. What will be the output of the program?
public class Test { public static void main (String [] args) { final Foo f = new Foo(); Thread t = new Thread(new Runnable() { public void run() { f.doStuff(); } }); Thread g = new Thread() { public void run() { f.doStuff(); } }; t.start(); g.start(); } } class Foo { int x = 5; public void doStuff() { if (x < 10) { // nothing to do try { wait(); } catch(InterruptedException ex) { } } else { System.out.println("x is " + x++); if (x >= 10) { notify(); } } } }
Options- A. The code will not compile because of an error on notify(); of class Foo.
- B. The code will not compile because of some other error in class Test.
- C. An exception occurs at runtime.
- D. It prints "x is 5 x is 6". Discuss
Correct Answer: An exception occurs at runtime.
Explanation:
C is correct because the thread does not own the lock of the object it invokes wait() on. If the method were synchronized, the code would run without exception.A, B are incorrect because the code compiles without errors.
D is incorrect because the exception is thrown before there is any output.
- 2. What will be the output of the program?
public class ThreadDemo { private int count = 1; public synchronized void doSomething() { for (int i = 0; i < 10; i++) System.out.println(count++); } public static void main(String[] args) { ThreadDemo demo = new ThreadDemo(); Thread a1 = new A(demo); Thread a2 = new A(demo); a1.start(); a2.start(); } } class A extends Thread { ThreadDemo demo; public A(ThreadDemo td) { demo = td; } public void run() { demo.doSomething(); } }
Options- A. It will print the numbers 0 to 19 sequentially
- B. It will print the numbers 1 to 20 sequentially
- C. It will print the numbers 1 to 20, but the order cannot be determined
- D. The code will not compile. Discuss
Correct Answer: It will print the numbers 1 to 20 sequentially
Explanation:
You have two different threads that share one reference to a common object.The updating and output takes place inside synchronized code.
One thread will run to completion printing the numbers 1-10.
The second thread will then run to completion printing the numbers 11-20.
- 3. What will be the output of the program?
public class SyncTest { public static void main (String [] args) { Thread t = new Thread() { Foo f = new Foo(); public void run() { f.increase(20); } }; t.start(); } } class Foo { private int data = 23; public void increase(int amt) { int x = data; data = x + amt; } }and assuming that data must be protected from corruption, what?if anything?can you add to the preceding code to ensure the integrity of data?
Options- A. Synchronize the run method.
- B. Wrap a synchronize(this) around the call to f.increase().
- C. The existing code will cause a runtime exception.
- D. Synchronize the increase() method Discuss
Correct Answer: Synchronize the increase() method
Explanation:
Option D is correct because synchronizing the code that actually does the increase will protect the code from being accessed by more than one thread at a time.Option A is incorrect because synchronizing the run() method would stop other threads from running the run() method (a bad idea) but still would not prevent other threads with other runnables from accessing the increase() method.
Option B is incorrect for virtually the same reason as A?synchronizing the code that calls the increase() method does not prevent other code from calling the increase() method.
- 4. What will be the output of the program?
public class WaitTest { public static void main(String [] args) { System.out.print("1 "); synchronized(args) { System.out.print("2 "); try { args.wait(); /* Line 11 */ } catch(InterruptedException e){ } } System.out.print("3 "); } }
Options- A. It fails to compile because the IllegalMonitorStateException of wait() is not dealt with in line 11.
- B. 1 2 3
- C. 1 3
- D. 1 2 Discuss
Correct Answer: 1 2
Explanation:
1 and 2 will be printed, but there will be no return from the wait call because no other thread will notify the main thread, so 3 will never be printed. The program is essentially frozen at line 11.A is incorrect; IllegalMonitorStateException is an unchecked exception so it doesn't have to be dealt with explicitly.
B and C are incorrect; 3 will never be printed, since this program will never terminate because it will wait forever.
- 5. What will be the output of the program?
public class Q126 implements Runnable { private int x; private int y; public static void main(String [] args) { Q126 that = new Q126(); (new Thread(that)).start( ); /* Line 8 */ (new Thread(that)).start( ); /* Line 9 */ } public synchronized void run( ) /* Line 11 */ { for (;;) /* Line 13 */ { x++; y++; System.out.println("x = " + x + "y = " + y); } } }
Options- A. An error at line 11 causes compilation to fail
- B. Errors at lines 8 and 9 cause compilation to fail.
- C. The program prints pairs of values for x and y that might not always be the same on the same line (for example, "x=2, y=1")
- D. The program prints pairs of values for x and y that are always the same on the same line (for example, "x=1, y=1". In addition, each value appears once (for example, "x=1, y=1" followed by "x=2, y=2") Discuss
Correct Answer: The program prints pairs of values for x and y that are always the same on the same line (for example, "x=1, y=1". In addition, each value appears once (for example, "x=1, y=1" followed by "x=2, y=2")
Explanation:
The synchronized code is the key to answering this question. Because x and y are both incremented inside the synchronized method they are always incremented together. Also keep in mind that the two threads share the same reference to the Q126 object.Also note that because of the infinite loop at line 13, only one thread ever gets to execute.
- 6. What will be the output of the program?
class MyThread extends Thread { public static void main(String [] args) { MyThread t = new MyThread(); t.start(); System.out.print("one. "); t.start(); System.out.print("two. "); } public void run() { System.out.print("Thread "); } }
Options- A. Compilation fails
- B. An exception occurs at runtime.
- C. It prints "Thread one. Thread two."
- D. The output cannot be determined. Discuss
Correct Answer: An exception occurs at runtime.
Explanation:
When the start() method is attempted a second time on a single Thread object, the method will throw an IllegalThreadStateException (you will not need to know this exception name for the exam). Even if the thread has finished running, it is still illegal to call start() again.- 7. 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:- 8. 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.- 9. 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.
- 10. 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
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- 1. What will be the output of the program?