Java "params" in method signature?
Asked Answered
S

3

116

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 parameter looks like an array but allows everyone using the method to pass as many parameters of that type as the caller wants.

I'm fairly sure Java supports similar behaviour, but I cant find out how to do it.

Sowens answered 6/2, 2009 at 10:4 Comment(0)
D
207

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).

Demean answered 6/2, 2009 at 10:4 Comment(2)
Thanks I bizzarrly found this out myself while I was looking up something else, and was coming here to answer the question myself.Sowens
Note that the ellipse is three periods (correctly typed in the answer as ...) and not an actually typed ellipse …Effete
P
11

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.

Planer answered 6/2, 2009 at 10:7 Comment(0)
B
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 
Branching answered 31/3, 2016 at 12:46 Comment(0)

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