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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.
More questions
- 1. Which of the following will directly stop the execution of a Thread?
Options- A. wait()
- B. notify()
- C. notifyall()
- D. exits synchronized code Discuss
Correct Answer: wait()
Explanation:
Option A is correct. wait() causes the current thread to wait until another thread invokes the notify() method or the notifyAll() method for this object.Option B is wrong. notify() - wakes up a single thread that is waiting on this object's monitor.
Option C is wrong. notifyAll() - wakes up all threads that are waiting on this object's monitor.
Option D is wrong. Typically, releasing a lock means the thread holding the lock (in other words, the thread currently in the synchronized method) exits the synchronized method. At that point, the lock is free until some other thread enters a synchronized method on that object. Does entering/exiting synchronized code mean that the thread execution stops? Not necessarily because the thread can still run code that is not synchronized. I think the word directly in the question gives us a clue. Exiting synchronized code does not directly stop the execution of a thread.
- 2. What will be the output of the program?
public class Test107 implements Runnable { private int x; private int y; public static void main(String args[]) { Test107 that = new Test107(); (new Thread(that)).start(); (new Thread(that)).start(); } public synchronized void run() { for(int i = 0; i < 10; i++) { x++; y++; System.out.println("x = " + x + ", y = " + y); /* Line 17 */ } } }
Options- A. Compilation error.
- B. Will print in this order: x = 1 y = 1 x = 2 y = 2 x = 3 y = 3 x = 4 y = 4 x = 5 y = 5... but the output will be produced by both threads running simultaneously.
- C. Will print in this order: x = 1 y = 1 x = 2 y = 2 x = 3 y = 3 x = 4 y = 4 x = 5 y = 5... but the output will be produced by first one thread then the other. This is guaranteed by the synchronised code.
- D. Will print in this order x = 1 y = 2 x = 3 y = 4 x = 5 y = 6 x = 7 y = 8... Discuss
Correct Answer: Will print in this order: x = 1 y = 1 x = 2 y = 2 x = 3 y = 3 x = 4 y = 4 x = 5 y = 5... but the output will be produced by first one thread then the other. This is guaranteed by the synchronised code.
Explanation:
Both threads are operating on the same instance variables. Because the code is synchronized the first thread will complete before the second thread begins. Modify line 17 to print the thread names:System.out.println(Thread.currentThread().getName() + " x = " + x + ", y = " + y);
- 3. What will be the output of the program?
String x = new String("xyz"); String y = "abc"; x = x + y;How many String objects have been created?
Options- A. 2
- B. 3
- C. 4
- D. 5 Discuss
Correct Answer: 4
Explanation:
Line 1 creates two, one referred to by x and the lost String "xyz". Line 2 creates one (for a total of three). Line 3 creates one more (for a total of four), the concatenated String referred to by x with a value of "xyzabc".- 4. What will be the output of the program?
public class SqrtExample { public static void main(String [] args) { double value = -9.0; System.out.println( Math.sqrt(value)); } }
Options- A. 3.0
- B. -3.0
- C. NaN
- D. Compilation fails. Discuss
Correct Answer: NaN
Explanation:
The sqrt() method returns NaN (not a number) when it's argument is less than zero.- 5. Which two are valid constructors for Thread?
- Thread(Runnable r, String name)
- Thread()
- Thread(int priority)
- Thread(Runnable r, ThreadGroup g)
- Thread(Runnable r, int priority)
Options- A. 1 and 3
- B. 2 and 4
- C. 1 and 2
- D. 2 and 5 Discuss
Correct Answer: 1 and 2
Explanation:
(1) and (2) are both valid constructors for Thread.(3), (4), and (5) are not legal Thread constructors, although (4) is close. If you reverse the arguments in (4), you'd have a valid constructor.
- 6. Which collection class allows you to associate its elements with key values, and allows you to retrieve objects in FIFO (first-in, first-out) sequence?
Options- A. java.util.ArrayList
- B. java.util.LinkedHashMap
- C. java.util.HashMap
- D. java.util.TreeMap Discuss
Correct Answer: java.util.LinkedHashMap
Explanation:
LinkedHashMap is the collection class used for caching purposes. FIFO is another way to indicate caching behavior. To retrieve LinkedHashMap elements in cached order, use the values() method and iterate over the resultant collection.- 7. What will be the output of the program?
String x = "xyz"; x.toUpperCase(); /* Line 2 */ String y = x.replace('Y', 'y'); y = y + "abc"; System.out.println(y);
Options- A. abcXyZ
- B. abcxyz
- C. xyzabc
- D. XyZabc Discuss
Correct Answer: xyzabc
Explanation:
Line 2 creates a new String object with the value "XYZ", but this new object is immediately lost because there is no reference to it. Line 3 creates a new String object referenced by y. This new String object has the value "xyz" because there was no "Y" in the String object referred to by x. Line 4 creates a new String object, appends "abc" to the value "xyz", and refers y to the result.- 8. What will be the output of the program?
public class SwitchTest { public static void main(String[] args) { System.out.println("value =" + switchIt(4)); } public static int switchIt(int x) { int j = 1; switch (x) { case l: j++; case 2: j++; case 3: j++; case 4: j++; case 5: j++; default: j++; } return j + x; } }
Options- A. value = 2
- B. value = 4
- C. value = 6
- D. value = 8 Discuss
Correct Answer: value = 8
Explanation:
Because there are no break statements, once the desired result is found, the program continues though each of the remaining options.- 9. What will be the output of the program?
public class If2 { static boolean b1, b2; public static void main(String [] args) { int x = 0; if ( !b1 ) /* Line 7 */ { if ( !b2 ) /* Line 9 */ { b1 = true; x++; if ( 5 > 6 ) { x++; } if ( !b1 ) x = x + 10; else if ( b2 = true ) /* Line 19 */ x = x + 100; else if ( b1 | b2 ) /* Line 21 */ x = x + 1000; } } System.out.println(x); } }
Options- A. 0
- B. 1
- C. 101
- D. 111 Discuss
Correct Answer: 101
Explanation:
As instance variables, b1 and b2 are initialized to false. The if tests on lines 7 and 9 are successful so b1 is set to true and x is incremented. The next if test to succeed is on line 19 (note that the code is not testing to see if b2 is true, it is setting b2 to be true). Since line 19 was successful, subsequent else-if's (line 21) will be skipped.- 10. What will be the output of the program?
public class A { void A() /* Line 3 */ { System.out.println("Class A"); } public static void main(String[] args) { new A(); } }
Options- A. Class A
- B. Compilation fails.
- C. An exception is thrown at line 3.
- D. The code executes with no output. Discuss
Correct Answer: The code executes with no output.
Explanation:
Option D is correct. The specification at line 3 is for a method and not a constructor and this method is never called therefore there is no output. The constructor that is called is the default constructor.
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