How do I create a method that accept varargs in Java?

Varargs (variable arguments)

is a new feature in Java 1.5 which allows us to pass multiple values in a single variable name when calling a method. Of course, it can be done easily using array but the varargs add another power to the language.

The varargs can be created by using three periods (...) after the parameter type. If a method accept others parameter than the varargs, the varargs parameter should be the last parameter to the method. And please be aware that overloading a varargs method can make harder to figure out which method is called in the code.

package org.kodejava.lang;

import java.util.Arrays;

public class VarArgsExample {
    public static void main(String[] args) {
        VarArgsExample e = new VarArgsExample();
        e.printParams(1, 2, 3);
        e.printParams(10, 20, 30, 40, 50);
        e.printParams(100, 200, 300, 400, 500);
    }

    public void printParams(int... numbers) {
        System.out.println(Arrays.toString(numbers));
    }
}

Running the code snippet give you the following output:

[1, 2, 3]
[10, 20, 30, 40, 50]
[100, 200, 300, 400, 500]

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Wayan

Founder at Kode Java

A programmer, runner and recreational diver, lives on the island of Bali, Indonesia. Programming in Java, Spring Framework, Hibernate / JPA, MySQL. To support me working on this project, buy me a cup of coffee, every little bit helps. Thank you.

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