Java “params” in method signature?

Question

In C#, if you want a method to have an indeterminate number of parameters, you can make the final parameter in the method signature a params so that the method

Answer 1

In Java it's called varargs, and the syntax looks like a regular parameter, but with an ellipsis ("...") after the type:

public void foo(Object... bar) {
    for (Object baz : bar) {
        System.out.println(baz.toString());
    }
}

The vararg parameter must always be the last parameter in the method signature, and is accessed as if you received an array of that type (e.g. Object[] in this case).

Answer 2

This will do the trick in Java

public void foo(String parameter, Object... arguments);

You have to add three points ... and the varagr parameter must be the last in the method's signature.

Answer 3

As it is written on previous answers, it is varargs and declared with ellipsis (...)

Moreover, you can either pass the value types and/or reference types or both mixed (google Autoboxing). Additionally you can use the method parameter as an array as shown with the printArgsAlternate method down below.

Demo Code

public class VarargsDemo {

    public static void main(String[] args) {
        printArgs(3, true, "Hello!", new Boolean(true), new Double(25.3), 'a', new Character('X'));
        printArgsAlternate(3, true, "Hello!", new Boolean(true), new Double(25.3), 'a', new Character('X'));
    }

    private static void printArgs(Object... arguments) {
        System.out.print("Arguments: ");
        for(Object o : arguments)
            System.out.print(o + " ");

        System.out.println();
    }

    private static void printArgsAlternate(Object... arguments) {
        System.out.print("Arguments: ");

        for(int i = 0; i < arguments.length; i++)
            System.out.print(arguments[i] + " ");

        System.out.println();
    }

}

Output

Arguments: 3 true Hello! true 25.3 a X 
Arguments: 3 true Hello! true 25.3 a X