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- Question
Which of the following are Java reserved words?
- run
- import
- default
- implement
Options- A. 1 and 2
- B. 2 and 3
- C. 3 and 4
- D. 2 and 4
- Correct Answer
- 2 and 3
Explanation(2) - This is a Java keyword(3) - This is a Java keyword
(1) - Is incorrect because although it is a method of Thread/Runnable it is not a keyword
(4) - This is not a Java keyword the keyword is implements
More questions
- 1. 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.
- 2. 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.
- 3. 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.
- 4. 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
- 5. 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.
- 6. 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".- 7. 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".- 8. What will be the output of the program?
public class ObjComp { public static void main(String [] args ) { int result = 0; ObjComp oc = new ObjComp(); Object o = oc; if (o == oc) result = 1; if (o != oc) result = result + 10; if (o.equals(oc) ) result = result + 100; if (oc.equals(o) ) result = result + 1000; System.out.println("result = " + result); } }
Options- A. 1
- B. 10
- C. 101
- D. 1101 Discuss
Correct Answer: 1101
Explanation:
Even though o and oc are reference variables of different types, they are both referring to the same object. This means that == will resolve to true and that the default equals() method will also resolve to true.- 9. What will be the output of the program?
String d = "bookkeeper"; d.substring(1,7); d = "w" + d; d.append("woo"); /* Line 4 */ System.out.println(d);
Options- A. wookkeewoo
- B. wbookkeeper
- C. wbookkeewoo
- D. Compilation fails. Discuss
Correct Answer: Compilation fails.
Explanation:
In line 4 the code calls a StringBuffer method, append() on a String object.- 10. What will be the output of the program?
interface Count { short counter = 0; void countUp(); } public class TestCount implements Count { public static void main(String [] args) { TestCount t = new TestCount(); t.countUp(); } public void countUp() { for (int x = 6; x>counter; x--, ++counter) /* Line 14 */ { System.out.print(" " + counter); } } }
Options- A. 0 1 2
- B. 1 2 3
- C. 0 1 2 3
- D. 1 2 3 4
- E. Compilation fails Discuss
Correct Answer: Compilation fails
Explanation:
The code will not compile because the variable counter is an interface variable that is by default final static. The compiler will complain at line 14 when the code attempts to increment counter.
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- 1. Which three statements are true?