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In Java, varargs is a type of variable that is declared within a method's parameters. It stands for variable arguments. A method parameter declared as the varargs type can be passed a variable number of arguments, i.e. zero or more number of arguments.

A point to remember - A varargs parameter can only be declared within a method's parameters.

• How to declare a varargs parameter


A varargs parameter is declared by specifying -
• Type of varargs parameter, followed by
• Three dots(...), followed by
• Name of the varargs parameter.



For example - let u s declare a method, meta(), that takes a varargs argument of type int in its parameters, which means that this method can be passed zero or more number of arguments of type int.



public void meta(int... i)








Note :


A varargs parameter can be a primitive type or an object type.






• Example of varargs




//Java - Example of varargs

class A
{

public void meth(int... a)
{
	System.out.println("Method meth is called");
}


public static void main(String... ar)
{
	A ob= new A();
	ob.meth(); 	//Caling meth() with zero argument
	ob.meth(1);	//Caling meth() with one argument
	ob.meth(1,2,3);	//Caling meth() with multiple arguments
}
}





Output is




Method meth is called
Method meth is called
Method meth is called


The meth() method is created with a varargs parameter of type int and we have successfully called this method with zero, one or multiple arguments.




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• A method can have a mix of non-varargs parameters and varargs parameters.


A method can have a mix of non-varargs parameters and varargs parameter declared in its signature. In such case, varargs must be the last parameter in the method's signature.



//Java - Example of varargs

class A
{

public void display(char c, int a, long... l)
{
	System.out.println("The value of a : " + a);
	System.out.println("The value of c : " + ch);
	System.out.println("The value of l : " + l);
}


public void display(char c, int... a, long l)
{
	System.out.println("The value of a : " + a);
	System.out.println("The value of c : " + ch);
	System.out.println("The value of l : " + l);
}
}




Output




var.java:12: error: ')' expected
public void display(char c, int... a, long l)
                                    ^
var.java:12: error: ';' expected
public void display(char c, int... a, long l)
                                            ^
2 errors


The above program throws a compile error because a varargs parameter wasn't declared as the last parameter in the method's signature.




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• A method cannot have more than one varargs parameter


A method cannot have more than one varargs parameter declared in its signature. Hence, the code below leads to a compile error.



//Java - Example of varargs


class A
{

public void show(int... a, char... ch)
{
	System.out.println("The value of a :" + a);
	System.out.println("The value of c :" + ch);
}

}




Output




var.java:4: error: ')' expected
public void show(int... a, char... ch)
                         ^
var.java:4: error: <identifier> expected
public void show(int... a, char... ch)
                               ^
var.java:4: error: <identifier> expected
public void show(int... a, char... ch)
                                     ^
3 errors


The above program throws a compile error because a method is not allowed to have multiple varargs parameters.






• Reading variable number of values passed to a method with varargs argument.

A varargs parameter is a variable-length array that holds zero or more values of same type. Hence, just like an array, a varargs parameter also has a length variable which gives the total number of elements contained in it.

//Java - Example of varargs

class A
{
public void meth(int... a)
{
	if(a.length==0)
	System.out.println("number of parameters passed : "+ a.length);

	if(a.length>0)
	{
		System.out.println("number of parameters passed : "+ a.length);

		for(int i=0;i<a.length;i++)	
		System.out.println(a[i]);
	}
}


public static void main(String... ar)
{
	A ob= new A();
	ob.meth();		//calling method() with zero parameter
	ob.meth(1);		//calling method() with one parameter
	ob.meth(1,2,3);  	//calling method() with more than one parameter
}
}

Output is

number of parameters passed : 0
number of parameters passed : 1
1
number of parameters passed : 3
1
2
3

We have called method meth() with zero, one and multiple arguments. These arguments are read using the length variable of varargs parameter.

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