VarArgs in java or Variable Arguments Tutorial with Examples

Table of Contents

Study and learn Interview MCQ Questions and Answers on VarArgs in java or Variable Arguments in Java. Attend job interviews easily with these Multiple Choice Questions

Java Varargs refer to Variable Arguments. Java Varargs are applicable only for Constructors and Methods. Let us know more in Java Tutorial.

VarArgs in java or Variable Arguments Explained

Java Varargs allow writing constructors or methods that accept any unknown (arbitrary) number of arguments. The number of arguments may be zero also. These Java Variable Arguments successfully implement Method Overloading or Constructor Overloading by maintaining unique method signatures.

You can make any Variable of Pri-mitive or Object type, a Variable Arguments (Varargs) type. To do this, you need to add Three Dots (…) next to the Data-Type within the Parameters List of a method or constructor.

A Java Vararg variable is treated as an Array. A Vararg variable can come only in the last position in a parameter list. In a single Method or Constructor, only one Varargs type variable defined.

Java Varargs in a Constructor

The syntax is as follows.

CONSTRUCTOR_NAME(TYPE... VARIABLE_NAME)
{
  //CODE for initialization
}

Check the below example that implements Varargs inside a Constructor’s Parameter List.

public class VarargsExample
{
  VarargsExample(int... a)
  {
    if(a.length == 0)
      System.out.println("Zero Arguments");
    for(int i=0; i<a.length; i++)
      System.out.print(a[i] + ",");
    System.out.println();
  }

  VarargsExample(String name, int...a)
  {
    System.out.println("Name= " + name);
    for(int i=0; i<a.length; i++)
      System.out.print(a[i] + ",");
  }

  public static void main(String[] args)
  {
    VarargsExample obj = new VarargsExample();
    VarargsExample ve = new VarargsExample(5,6,7);
    VarargsExample ve2 = new VarargsExample("CANADA", 1,2);
  }
}
//OUTPUT
Zero Arguments

5,6,7,
Name= CANADA
1,2,

Notice that we have not maintained a Zero-Argument constructor in the above example. Even then, the constructor with int…a arguments is invoked for new VarargsExample() call. If you put a constructor with zero arguments, it will be invoked instead of Variable Arguments type constructor if it is the first constructor matched for zero arguments.

Let us check another example that shows incompatible constructor overloading using Varargs.

public class VarargsExample
{
  VarargsExample(boolean...a)
  {

  }
  VarargsExample(int... a)
  {

  }
  public static void main(String[] args)
  {
    VarargsExample ve = new VarargsExample(); //Error
  }

}
//OUTPUT
Compiler error

The above example produces an error “The Constructor is Ambiguous“. Because int and boolean are incompatible or inconvertible data types. If we replace boolean with a number data type like float, double, long or byte, the program compiles and gives the output.

Note: While choosing a Vararg constructor, the compiler selects the one with a small-sized data type first. So CONSTRUCTOR(byte…a) is selected before CONSTRUCTOR(int…a).

Java Varargs in a Method

Just like in a Constructor with Varargs, Java methods too can have Varargs. Only one Vararg type variable is allowed in a method that too at the end of the parameters list.

The syntax is as follows.

METHOD_NAME(TYPE... VARIABLE_NAME)
{
  //CODE for initialization
}

Check the below example that implements Varargs inside a Method’s Parameter List.

public class VarargsMethodExample
{
  void show(String...names)
  {

  }
  void show(int...codes)
  {

  }
  void modify(String operator, long...transmission)
  {

  }
  String report(float var, int...codes)
  {
    String a = "";
    for(int i=0; i<codes.length; i++)
    {
      a = a + codes[i] + "," ;
    }
    return a;
  }

  public static void main(String[] args)
  {
    VarargsMethodExample vme = new VarargsMethodExample();
    String rep = vme.report(0.0f, 2,3,4);
    System.out.println("Report= " + rep);
  }
}
//OUTPUT
Report= 2,3,4,

In the above example, we have used Java Varargs in the methods show(), modify() and report(). Method “show” is overloaded. Method “report” even returns a value.

[WpProQuiz 163]

VarArgs in JAVA Interview MCQ

1) Java Varargs are applicable only for ___.

A) Constructors

B) Methods

C) Both Constructors and Methods

D) None

Answer [=] C

Explanation:

Only constructors and methods accept arguments. So, Java Varargs are applicable only to these.

2) A Java-Vararg is nothing but ____.

A) Variable number of arguments

B) Variable type of arguments

C) Variable size of arguments

D) All

Answer [=] A

Explanation:

A Java Vararg represents simply a variable number of arguments.

3) A Java vararg is a ____.

A) Method

B) Constructor

C) Variable

D) All

Answer [=] C

Explanation:

A Java-Vararg represents a variable of a particular type.

4) A Java-Vararg can be of any type like pri-mitive or object type. State TRUE or FALSE.

A) TRUE

B) FALSE

Answer [=] A

Explanation:

Yes. A Vararg can be a non-primitive or object type also along with a pri-mitive type like byte, short, int, long, float, double etc.

5) A Java Vararg or Variable Argument can come at any position in a method or constructor. State TRUE or FALSE.

A) FALSE

B) TRUE

Answer [=] B

Explanation:

A Vararg can come only as the last argument in the list of arguments of a constructor or method.

6) What is the output of the below java program with varargs?

public class Varargs1
{
  static void displayStudents(String... stu)
  {
    for(String s: stu)
      System.out.print(s + " ");
  }
  public static void main(String args[])
  {
    displayStudents("Bean", "Atkinson", "Milton");
  }
}

A) Bean Bean Bean

B) null null null

C) Bean Atkinson Milton

D) Compiler error

Answer [=] C

Explanation:

In the above example, the Vararg variable “stu” is of String array type. So, we have used a FOR loop to go through all the elements. We also use a NORMAL FOR loop with an index starting from 0.

7) What is the output of the below Java program with Variable arguments?

public class Varargs2
{
  void attendance(String... allStu)
  {
    System.out.println("Attended: " + allStu.length);
  }
  void attendance(boolean... all)
  {
    System.out.println("Attended: " + all.length);
  }
  public static void main(String args[])
  {
    new Varargs2().attendance();
  }
}

Attended: 0

Attended: 0
Attended: 0

C) No Output

D) Compiler Error

Answer [=] D

Explanation:

Observe that there is no default constructor with 0 arguments. When no argument is passed, the compiler can not choose which attendance() method to choose. So it gives an error.

8) Which is the error thrown when two methods with varargs look the same to the compiler?

A) The method is ambiguous

B) The method is difficult to choose

C) The method signature is not correct

D) None

Answer [=] A

Explanation:

The compiler simply gives an error when two methods look the same to the compiler for calls at compile time.

9) What is the output of the below Java program with Varargs?

public class Varargs3
{
  Varargs3(int... dates)
  {
    System.out.println("Inside Varargs(int...)");
  }
  Varargs3(boolean... yesno)
  {
    System.out.println("Inside Varargs(float...)");
  }
  public static void main(String[] args)
  {
    new Varargs3();
  }
}

Inside Varargs(int...)

Inside Varargs(boolean...)

Inside Varargs(int...)
Inside Varargs(boolean...)

D) Compiler error

Answer [=] D

Explanation:

As the two constructors have the same method signature with zero arguments, the compiler sees those as the same. So, it throws an error saying the Constructor is ambiguous.

10) What is the output of the below Java program with Varargs?

public class Varargs4
{
  Varargs4(int... carnums)
  {
    System.out.println("Inside Varargs(int...)");
  }
  Varargs4(float... prices)
  {
    System.out.println("Inside Varargs(float...)");
  }
  public static void main(String[] args)
  {
    new Varargs4();
  }
}

Inside Varargs(int...)

Inside Varargs(int...)
Inside Varargs(float...)

C) No output

D) Compiler error

Answer [=] A

Explanation:

In the above example, both constructors are overloading one another. In such cases, the compiler chooses the Constructor or Method with a lower sized data type as an argument.

11) What is the output of the below code snippet?

public class Varargs5
{
  Varargs5(int...weights, boolean yesno)
  {
    System.out.println("AMAZON");
  }
  public static void main(String[] args)
  {
   //new Varargs5(20, true);
  }
}

A) No output

B) Error: The variable argument type int of the method Varargs5 must be the last parameter

C) Error: Varargs do not allow other data types

D) None

Answer [=] B

Explanation:

Yes. The compiler throws errors “Unresolved Compilation Problem” and “The variable argument type of the method or constructor must be the last parameter”

12) Which is the operator used to represent a Vararg type in a method or constructor in Java?

A) One Dot (.)

B) Two Dots (..)

C) Three Dots (…)

D) DOT DOT DOT COMMA (…,)

Answer [=] C

Explanation:

The data type (pri-mitive or object) immediately followed by three dots and a variable name separated by a space create a Variable Argument.

13) How many maximum numbers of Varargs or Variable-Arguments can be there in a method or a constructor in Java?

A) 1

B) 2

C) 8

D) 16

Answer [=] A

Explanation:

Yes, only one. Because a VARARG can be present only as the last parameter or argument.

14) What is the maximum number of methods or constructors with Varargs in a single Java class?

A) 1

B) 2

C) 8

D) There is no limit

Answer [=] D

Explanation:

Yes. There is no limit. There can be any number of methods or constructors with one Vararg per each method or constructor.

For More Free online Quizzes of JAVA PROGRAMMING Test online Click here

For More Free online Test of previous years Question Papers of Aptitude Click here

onlineexamguide