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- Question
In the given program, how many lines of output will be produced?
public class Test { public static void main(String [] args) { int [] [] [] x = new int [3] [] []; int i, j; x[0] = new int[4][]; x[1] = new int[2][]; x[2] = new int[5][]; for (i = 0; i < x.length; i++) { for (j = 0; j < x[i].length; j++) { x[i][j] = new int [i + j + 1]; System.out.println("size = " + x[i][j].length); } } } }
Options- A. 7
- B. 9
- C. 11
- D. 13
- E. Compilation fails
- Correct Answer
- 11
ExplanationThe loops use the array sizes (length).It produces 11 lines of output as given below.
D:\Java>javac Test.java D:\Java>java Test size = 1 size = 2 size = 3 size = 4 size = 2 size = 3 size = 3 size = 4 size = 5 size = 6 size = 7
Therefore, 11 is the answer.
More questions
- 1. 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.- 2. 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.
- 3. 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.- 4. 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.
- 5. 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.
- 6. 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.
- 7. 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
- 8. 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.
- 9. What will be the output of the program?
class Tree { } class Pine extends Tree { } class Oak extends Tree { } public class Forest1 { public static void main (String [] args) { Tree tree = new Pine(); if( tree instanceof Pine ) System.out.println ("Pine"); else if( tree instanceof Tree ) System.out.println ("Tree"); else if( tree instanceof Oak ) System.out.println ( "Oak" ); else System.out.println ("Oops "); } }
Options- A. Pine
- B. Tree
- C. Forest
- D. Oops Discuss
Correct Answer: Pine
Explanation:
The program prints "Pine".- 10. What will be the output of the program?
public class StringRef { public static void main(String [] args) { String s1 = "abc"; String s2 = "def"; String s3 = s2; /* Line 7 */ s2 = "ghi"; System.out.println(s1 + s2 + s3); } }
Options- A. abcdefghi
- B. abcdefdef
- C. abcghidef
- D. abcghighi Discuss
Correct Answer: abcghidef
Explanation:
After line 7 executes, both s2 and s3 refer to a String object that contains the value "def". When line 8 executes, a new String object is created with the value "ghi", to which s2 refers. The reference variable s3 still refers to the (immutable) String object with the value "def".
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- 1. What will be the output of the program?