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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. Which is true about a method-local inner class?
Options- A. It must be marked final.
- B. It can be marked abstract.
- C. It can be marked public.
- D. It can be marked static. Discuss
Correct Answer: It can be marked abstract.
Explanation:
Option B is correct because a method-local inner class can be abstract, although it means a subclass of the inner class must be created if the abstract class is to be used (so an abstract method-local inner class is probably not useful).Option A is incorrect because a method-local inner class does not have to be declared final (although it is legal to do so).
C and D are incorrect because a method-local inner class cannot be made public (remember-you cannot mark any local variables as public), or static.
- 2. Given that all methods of class FileOutputStream, including close(), throw an IOException, which of these is true?
import java.io.*; public class MyProgram { public static void main(String args[]) { FileOutputStream out = null; try { out = new FileOutputStream("test.txt"); out.write(122); } catch(IOException io) { System.out.println("IO Error."); } finally { out.close(); } } }
Options- A. This program will compile successfully.
- B. This program fails to compile due to an error at line 4.
- C. This program fails to compile due to an error at line 6.
- D. This program fails to compile due to an error at line 18. Discuss
Correct Answer: This program fails to compile due to an error at line 18.
Explanation:
Any method (in this case, the main() method) that throws a checked exception (in this case, out.close() ) must be called within a try clause, or the method must declare that it throws the exception. Either main() must declare that it throws an exception, or the call to out.close() in the finally block must fall inside a (in this case nested) try-catch block.- 3. What will be the output of the program?
public class Test { public static void main(String [] args) { int I = 1; do while ( I < 1 ) System.out.print("I is " + I); while ( I > 1 ) ; } }
Options- A. I is 1
- B. I is 1 I is 1
- C. No output is produced.
- D. Compilation error Discuss
Correct Answer: No output is produced.
Explanation:
There are two different looping constructs in this problem. The first is a do-while loop and the second is a while loop, nested inside the do-while. The body of the do-while is only a single statement-brackets are not needed. You are assured that the while expression will be evaluated at least once, followed by an evaluation of the do-while expression. Both expressions are false and no output is produced.- 4. What will be the output of the program?
class MyThread extends Thread { public static void main(String [] args) { MyThread t = new MyThread(); Thread x = new Thread(t); x.start(); /* Line 7 */ } public void run() { for(int i = 0; i < 3; ++i) { System.out.print(i + ".."); } } }
Options- A. Compilation fails.
- B. 1..2..3..
- C. 0..1..2..3..
- D. 0..1..2.. Discuss
Correct Answer: 0..1..2..
Explanation:
The thread MyThread will start and loop three times (from 0 to 2).Option A is incorrect because the Thread class implements the Runnable interface; therefore, in line 7, Thread can take an object of type Thread as an argument in the constructor.
Option B and C are incorrect because the variable i in the for loop starts with a value of 0 and ends with a value of 2.
- 5. Which of the following is/are legal method declarations?
- protected abstract void m1();
- static final void m1(){}
- synchronized public final void m1() {}
- private native void m1();
Options- A. 1 and 3
- B. 2 and 4
- C. 1 only
- D. All of them are legal declarations. Discuss
Correct Answer: All of them are legal declarations.
Explanation:
All the given statements are legal declarations.- 6. Which three statements are true?
- f1 == f2
- f1 == f3
- f2 == f1[1]
- x == f1[0]
- f == f1[0]
import java.awt.Button; class CompareReference { public static void main(String [] args) { float f = 42.0f; float [] f1 = new float[2]; float [] f2 = new float[2]; float [] f3 = f1; long x = 42; f1[0] = 42.0f; } }
Options- A. 1, 2 and 3
- B. 2, 4 and 5
- C. 3, 4 and 5
- D. 1, 4 and 5 Discuss
Correct Answer: 2, 4 and 5
Explanation:
(2) is correct because the reference variables f1 and f3 refer to the same array object.(4) is correct because it is legal to compare integer and floating-point types.
(5) is correct because it is legal to compare a variable with an array element.
(3) is incorrect because f2 is an array object and f1[1] is an array element.
- 7. Which one create an anonymous inner class from within class Bar?
class Boo { Boo(String s) { } Boo() { } } class Bar extends Boo { Bar() { } Bar(String s) {super(s);} void zoo() { // insert code here } }
Options- A. Boo f = new Boo(24) { };
- B. Boo f = new Bar() { };
- C. Bar f = new Boo(String s) { };
- D. Boo f = new Boo.Bar(String s) { }; Discuss
Correct Answer: Boo f = new Bar() { };
Explanation:
Option B is correct because anonymous inner classes are no different from any other class when it comes to polymorphism. That means you are always allowed to declare a reference variable of the superclass type and have that reference variable refer to an instance of a subclass type, which in this case is an anonymous subclass of Bar. Since Bar is a subclass of Boo, it all works.Option A is incorrect because it passes an int to the Boo constructor, and there is no matching constructor in the Boo class.
Option C is incorrect because it violates the rules of polymorphism—you cannot refer to a superclass type using a reference variable declared as the subclass type. The superclass is not guaranteed to have everything the subclass has.
Option D uses incorrect syntax.
- 8. Which of the following code fragments inserted, will allow to compile?
public class Outer { public void someOuterMethod() { //Line 5 } public class Inner { } public static void main(String[] argv) { Outer ot = new Outer(); //Line 10 } }
Options- A. new Inner(); //At line 5
- B. new Inner(); //At line 10
- C. new ot.Inner(); //At line 10
- D. new Outer.Inner(); //At line 10 Discuss
Correct Answer: new Inner(); //At line 5
Explanation:
Option A compiles without problem.Option B gives error - non-static variable cannot be referenced from a static context.
Option C package ot does not exist.
Option D gives error - non-static variable cannot be referenced from a static context.
- 9. What will be the output of the program?
public class Test178 { public static void main(String[] args) { String s = "foo"; Object o = (Object)s; if (s.equals(o)) { System.out.print("AAA"); } else { System.out.print("BBB"); } if (o.equals(s)) { System.out.print("CCC"); } else { System.out.print("DDD"); } } }
Options- A. AAACCC
- B. AAADDD
- C. BBBCCC
- D. BBBDDD Discuss
Correct Answer: AAACCC
- 10. What will be the output of the program?
int i = (int) Math.random();
Options- A. i = 0
- B. i = 1
- C. value of i is undetermined
- D. Statement causes a compile error Discuss
Correct Answer: i = 0
Explanation:
Math.random() returns a double value greater than or equal to 0 and less than 1. Its value is stored to an int but as this is a narrowing conversion, a cast is needed to tell the compiler that you are aware that there may be a loss of precision.The value after the decimal point is lost when you cast a double to int and you are left with 0.
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