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- Question
What will be the output of the program?
int i = 1, j = 10; do { if(i > j) { break; } j--; } while (++i < 5); System.out.println("i = " + i + " and j = " + j);
Options- A. i = 6 and j = 5
- B. i = 5 and j = 5
- C. i = 6 and j = 4
- D. i = 5 and j = 6
- Correct Answer
- i = 5 and j = 6
ExplanationThis loop is a do-while loop, which always executes the code block within the block at least once, due to the testing condition being at the end of the loop, rather than at the beginning. This particular loop is exited prematurely if i becomes greater than j.The order is, test i against j, if bigger, it breaks from the loop, decrements j by one, and then tests the loop condition, where a pre-incremented by one i is tested for being lower than 5. The test is at the end of the loop, so i can reach the value of 5 before it fails. So it goes, start:
1, 10
2, 9
3, 8
4, 7
5, 6 loop condition fails.
More questions
- 1. Given that EOFException and FileNotFoundException are both subclasses of IOException, and further assuming this block of code is placed into a class, which statement is most true concerning this code?
System.out.print("Start "); try { System.out.print("Hello world"); throw new FileNotFoundException(); } System.out.print(" Catch Here "); /* Line 7 */ catch(EOFException e) { System.out.print("End of file exception"); } catch(FileNotFoundException e) { System.out.print("File not found"); }
Options- A. The code will not compile.
- B. Code output: Start Hello world File Not Found.
- C. Code output: Start Hello world End of file exception.
- D. Code output: Start Hello world Catch Here File not found. Discuss
Correct Answer: The code will not compile.
Explanation:
Line 7 will cause a compiler error. The only legal statements after try blocks are either catch or finally statements.Option B, C, and D are incorrect based on the program logic described above. If line 7 was removed, the code would compile and the correct answer would be Option B.
- 2. Which three piece of codes are equivalent to line 3?
- final int k = 4;
- public int k = 4;
- static int k = 4;
- abstract int k = 4;
- volatile int k = 4;
- protected int k = 4;
public interface Foo { int k = 4; /* Line 3 */ }
Options- A. 1, 2 and 3
- B. 2, 3 and 4
- C. 3, 4 and 5
- D. 4, 5 and 6 Discuss
Correct Answer: 1, 2 and 3
Explanation:
(1), (2) and (3) are correct. Interfaces can have constants, which are always implicitly public, static, and final. Interface constant declarations of public, static, and final are optional in any combination.- 3. After line 11 runs, how many objects are eligible for garbage collection?
class X2 { public X2 x; public static void main(String [] args) { X2 x2 = new X2(); /* Line 6 */ X2 x3 = new X2(); /* Line 7 */ x2.x = x3; x3.x = x2; x2 = new X2(); x3 = x2; /* Line 11 */ doComplexStuff(); } }
Options- A. 0
- B. 1
- C. 2
- D. 3 Discuss
Correct Answer: 2
Explanation:
This is an example of the islands of isolated objects. By the time line 11 has run, the objects instantiated in lines 6 and 7 are referring to each other, but no live thread can reach either of them.- 4. When is the Float object, created in line 3, eligible for garbage collection?
public Object m() { Object o = new Float(3.14F); Object [] oa = new Object[l]; oa[0] = o; /* Line 5 */ o = null; /* Line 6 */ oa[0] = null; /* Line 7 */ return o; /* Line 8 */ }
Options- A. just after line 5
- B. just after line 6
- C. just after line 7
- D. just after line 8 Discuss
Correct Answer: just after line 7
Explanation:
Option A is wrong. This simply copies the object reference into the array.Option B is wrong. The reference o is set to null, but, oa[0] still maintains the reference to the Float object.
Option C is correct. The thread of execution will then not have access to the object.
- 5. What is the prototype of the default constructor?
public class Test { }
Options- A. Test( )
- B. Test(void)
- C. public Test( )
- D. public Test(void) Discuss
Correct Answer: public Test( )
Explanation:
Option A and B are wrong because they use the default access modifier and the access modifier for the class is public (remember, the default constructor has the same access modifier as the class).Option D is wrong. The void makes the compiler think that this is a method specification - in fact if it were a method specification the compiler would spit it out.
- 6. Which is valid in a class that extends class A?
class A { protected int method1(int a, int b) { return 0; } }
Options- A. public int method1(int a, int b) {return 0; }
- B. private int method1(int a, int b) { return 0; }
- C. public short method1(int a, int b) { return 0; }
- D. static protected int method1(int a, int b) { return 0; } Discuss
Correct Answer: public int method1(int a, int b) {return 0; }
Explanation:
Option A is correct - because the class that extends A is just simply overriding method1.Option B is wrong - because it can't override as there are less access privileges in the subclass method1.
Option C is wrong - because to override it, the return type needs to be an integer. The different return type means that the method is not overriding but the same argument list means that the method is not overloading. Conflict - compile time error.
Option D is wrong - because you can't override a method and make it a class method i.e. using static.
- 7. After line 8 runs. how many objects are eligible for garbage collection?
public class X { public static void main(String [] args) { X x = new X(); X x2 = m1(x); /* Line 6 */ X x4 = new X(); x2 = x4; /* Line 8 */ doComplexStuff(); } static X m1(X mx) { mx = new X(); return mx; } }
Options- A. 0
- B. 1
- C. 2
- D. 3 Discuss
Correct Answer: 1
Explanation:
By the time line 8 has run, the only object without a reference is the one generated as a result of line 6. Remember that "Java is pass by value," so the reference variable x is not affected by the m1() method.- 8. 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.
- 9. 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);
- 10. 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".
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- 1. Given that EOFException and FileNotFoundException are both subclasses of IOException, and further assuming this block of code is placed into a class, which statement is most true concerning this code?