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- Question
What will be the output of the program?
class Super { public int i = 0; public Super(String text) /* Line 4 */ { i = 1; } } class Sub extends Super { public Sub(String text) { i = 2; } public static void main(String args[]) { Sub sub = new Sub("Hello"); System.out.println(sub.i); } }
Options- A. 0
- B. 1
- C. 2
- D. Compilation fails.
- Correct Answer
- Compilation fails.
ExplanationA default no-args constructor is not created because there is a constructor supplied that has an argument, line 4. Therefore the sub-class constructor must explicitly make a call to the super class constructor:
Declarations and Access Control problems
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- 1. What will be the output of the program?
class Super { public Integer getLength() { return new Integer(4); } } public class Sub extends Super { public Long getLength() { return new Long(5); } public static void main(String[] args) { Super sooper = new Super(); Sub sub = new Sub(); System.out.println( sooper.getLength().toString() + "," + sub.getLength().toString() ); } }
Options- A. 4, 4
- B. 4, 5
- C. 5, 4
- D. Compilation fails. Discuss
Correct Answer: Compilation fails.
Explanation:
Option D is correct, compilation fails - The return type of getLength( ) in the super class is an object of reference type Integer and the return type in the sub class is an object of reference type Long. In other words, it is not an override because of the change in the return type and it is also not an overload because the argument list has not changed.- 2. What will be the output of the program?
public class Test { public int aMethod() { static int i = 0; i++; return i; } public static void main(String args[]) { Test test = new Test(); test.aMethod(); int j = test.aMethod(); System.out.println(j); } }
Options- A. 0
- B. 1
- C. 2
- D. Compilation fails. Discuss
Correct Answer: Compilation fails.
Explanation:
Compilation failed because static was an illegal start of expression - method variables do not have a modifier (they are always considered local).- 3. What will be the output of the program?
public class ArrayTest { public static void main(String[ ] args) { float f1[ ], f2[ ]; f1 = new float[10]; f2 = f1; System.out.println("f2[0] = " + f2[0]); } }
Options- A. It prints f2[0] = 0.0
- B. It prints f2[0] = NaN
- C. An error at f2 = f1; causes compile to fail.
- D. It prints the garbage value. Discuss
Correct Answer: It prints f2[0] = 0.0
Explanation:
Option A is correct. When you create an array (f1 = new float[10];) the elements are initialises to the default values for the primitive data type (float in this case - 0.0), so f1 will contain 10 elements each with a value of 0.0. f2 has been declared but has not been initialised, it has the ability to reference or point to an array but as yet does not point to any array. f2 = f1; copies the reference (pointer/memory address) of f1 into f2 so now f2 points at the array pointed to by f1.This means that the values returned by f2 are the values returned by f1. Changes to f1 are also changes to f2 because both f1 and f2 point to the same array.
- 4. What will be the output of the program?
public class Test { public static void main(String args[]) { class Foo { public int i = 3; } Object o = (Object)new Foo(); Foo foo = (Foo)o; System.out.println("i = " + foo.i); } }
Options- A. i = 3
- B. Compilation fails.
- C. i = 5
- D. A ClassCastException will occur. Discuss
Correct Answer: i = 3
- 5. What will be the output of the program?
import java.util.*; public class NewTreeSet2 extends NewTreeSet { public static void main(String [] args) { NewTreeSet2 t = new NewTreeSet2(); t.count(); } } protected class NewTreeSet { void count() { for (int x = 0; x < 7; x++,x++ ) { System.out.print(" " + x); } } }
Options- A. 0 2 4
- B. 0 2 4 6
- C. Compilation fails at line 2
- D. Compilation fails at line 10 Discuss
Correct Answer: Compilation fails at line 10
Explanation:
Nonnested classes cannot be marked protected (or final for that matter), so the compiler will fail at protected class NewTreeSet.- 6. What will be the output of the program?
class Base { Base() { System.out.print("Base"); } } public class Alpha extends Base { public static void main(String[] args) { new Alpha(); /* Line 12 */ new Base(); /* Line 13 */ } }
Options- A. Base
- B. BaseBase
- C. Compilation fails
- D. The code runs with no output Discuss
Correct Answer: BaseBase
Explanation:
Option B is correct. It would be correct if the code had compiled, and the subclass Alpha had been saved in its own file. In this case Java supplies an implicit call from the sub-class constructor to the no-args constructor of the super-class therefore line 12 causes Base to be output. Line 13 also causes Base to be output.Option A is wrong. It would be correct if either the main class or the subclass had not been instantiated.
Option C is wrong. The code compiles.
Option D is wrong. There is output.
- 7. What will be the output of the program?
class A { final public int GetResult(int a, int b) { return 0; } } class B extends A { public int GetResult(int a, int b) {return 1; } } public class Test { public static void main(String args[]) { B b = new B(); System.out.println("x = " + b.GetResult(0, 1)); } }
Options- A. x = 0
- B. x = 1
- C. Compilation fails.
- D. An exception is thrown at runtime. Discuss
Correct Answer: Compilation fails.
Explanation:
The code doesn't compile because the method GetResult() in class A is final and so cannot be overridden.- 8. 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 Discuss
Correct Answer: i = 5 and j = 6
Explanation:
This 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.
- 9. What will be the output of the program?
boolean bool = true; if(bool = false) /* Line 2 */ { System.out.println("a"); } else if(bool) /* Line 6 */ { System.out.println("b"); } else if(!bool) /* Line 10 */ { System.out.println("c"); /* Line 12 */ } else { System.out.println("d"); }
Options- A. a
- B. b
- C. c
- D. d Discuss
Correct Answer: c
Explanation:
Look closely at line 2, is this an equality check (==) or an assignment (=). The condition at line 2 evaluates to false and also assigns false to bool. bool is now false so the condition at line 6 is not true. The condition at line 10 checks to see if bool is not true ( if !(bool == true) ), it isn't so line 12 is executed.- 10. 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.
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- 1. What will be the output of the program?