Java Questions & Answers – Interfaces – 1

This section of our 1000+ Java MCQs focuses on interfaces of Java Programming Language.

1. Which of these keywords is used to define interfaces in Java?
a) interface
b) Interface
c) intf
d) Intf
View Answer

Answer: a
Explanation: None.

2. Which of these can be used to fully abstract a class from its implementation?
a) Objects
b) Packages
c) Interfaces
d) None of the Mentioned
View Answer

Answer: c
Explanation: None.

3. Which of these access specifiers can be used for an interface?
a) Public
b) Protected
c) private
d) All of the mentioned
View Answer

Answer: a
Explanation: Access specifier of an interface is either public or no specifier. When no access specifier is used then default access specifier is used due to which interface is available only to other members of the package in which it is declared, when declared public it can be used by any code.
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4. Which of these keywords is used by a class to use an interface defined previously?
a) import
b) Import
c) implements
d) Implements
View Answer

Answer: c
Explanation: interface is inherited by a class using implements.

5. Which of the following is the correct way of implementing an interface salary by class manager?
a) class manager extends salary {}
b) class manager implements salary {}
c) class manager imports salary {}
d) none of the mentioned
View Answer

Answer: b
Explanation: None.
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6. Which of the following is an incorrect statement about packages?
a) Interfaces specifies what class must do but not how it does
b) Interfaces are specified public if they are to be accessed by any code in the program
c) All variables in interface are implicitly final and static
d) All variables are static and methods are public if interface is defined pubic
View Answer

Answer: d
Explanation: All methods and variables are implicitly public if interface is declared public.

7. What will be the output of the following Java program?

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  1.     interface calculate
  2.     {
  3.         void cal(int item);
  4.     }
  5.     class display implements calculate
  6.     {
  7.         int x;
  8.         public void cal(int item)
  9.         {
  10.             x = item * item;            
  11.         }
  12.     }
  13.     class interfaces
  14.     {
  15.         public static void main(String args[])
  16.         {
  17.             display arr = new display;
  18.             arr.x = 0;      
  19.             arr.cal(2);
  20.             System.out.print(arr.x);
  21.         }
  22.     }

a) 0
b) 2
c) 4
d) None of the mentioned
View Answer

Answer: c
Explanation: None.
Output:

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$ javac interfaces.java
$ java interfaces
4

8. What will be the output of the following Java program?

  1.     interface calculate
  2.     {
  3.         void cal(int item);
  4.     }
  5.     class displayA implements calculate
  6.     {
  7.         int x;
  8.         public void cal(int item)
  9.         {
  10.             x = item * item;            
  11.         }
  12.     }
  13.     class displayB implements calculate
  14.     {
  15.         int x;
  16.         public void cal(int item)
  17.         {
  18.             x = item / item;            
  19.         }
  20.     }
  21.     class interfaces 
  22.     {
  23.         public static void main(String args[])
  24.         {
  25.             displayA arr1 = new displayA;
  26.             displayB arr2 = new displayB;
  27.             arr1.x = 0;
  28.             arr2.x = 0;      
  29.             arr1.cal(2);
  30.             arr2.cal(2);
  31.             System.out.print(arr1.x + " " + arr2.x);
  32.         }
  33.     }

a) 0 0
b) 2 2
c) 4 1
d) 1 4
View Answer

Answer: c
Explanation: class displayA implements the interface calculate by doubling the value of item, where as class displayB implements the interface by dividing item by item, therefore variable x of class displayA stores 4 and variable x of class displayB stores 1.
Output:

$ javac interfaces.java
$ java interfaces
4 1

9. What will be the output of the following Java program?

  1. interface calculate 
  2. {
  3.             int VAR = 0;
  4.             void cal(int item);
  5. }
  6.         class display implements calculate 
  7.         {
  8.             int x;
  9.           public  void cal(int item)
  10.           {
  11.                 if (item<2)
  12.                     x = VAR;
  13.                 else
  14.                     x = item * item;            
  15.             }
  16.         }
  17.  class interfaces 
  18. {
  19.  
  20.             public static void main(String args[]) 
  21.             {
  22.                 display[] arr=new display[3];
  23.  
  24.                for(int i=0;i<3;i++)
  25.                arr[i]=new display();
  26.                arr[0].cal(0);    
  27.                arr[1].cal(1);
  28.                arr[2].cal(2);
  29.                System.out.print(arr[0].x+" " + arr[1].x + " " + arr[2].x);
  30.             }
  31. }

a) 0 1 2
b) 0 2 4
c) 0 0 4
d) 0 1 4
View Answer

Answer: c
Explanation: None.
output:

$ javac interfaces.java
$ java interfaces 
0 0 4

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Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He lives in Bangalore, and focuses on development of Linux Kernel, SAN Technologies, Advanced C, Data Structures & Alogrithms. Stay connected with him at LinkedIn.

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