What is the "..." operator used while receiving arrays in java method arguments

Question

public class Venus {
    public static void main(String[] args) {
        int [] x = {1,2,3};
        int y[] = {4,5,6};
        new Venus().go(x,y);
    }
    void go(int[]... z) { 
        /*
         *here where my doubt goes about how the "..." operator works
         */

    for(int[] a : z)
        System.out.print(a[0]);
}

}

output of this code snippet is "14"

[Varargs](https://docs.oracle.com/javase/1.5.0/docs/guide/language/varargs.html) — MadProgrammer, Dec 17 '17 at 09:19
https://stackoverflow.com/questions/766559/when-do-you-use-varargs-in-java — svarog, Dec 17 '17 at 09:19
Please consider reading a Java tutorial or textbook. For example: https://docs.oracle.com/javase/tutorial/java/javaOO/arguments.html — Stephen C, Dec 17 '17 at 15:09

score 0 · Accepted Answer · answered Dec 17 '17 at 09:22

0

... is the elipsis (varargs) operator. In short, it's a syntactic sugar that allows you to pass a variable number of arguments to the method using commas (,). Inside the method, the argument is treated as an array.

So int[]... z means that z is treated as an array of int[], or int[][].

answered Dec 17 '17 at 09:22

Mureinik

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If you think that this answer was helpful you might want to consider upvoting this answer to say "thank you" another way here at Stack Overlflow :-) – Przemysław Moskal Dec 17 '17 at 09:37

What is the "..." operator used while receiving arrays in java method arguments

1 Answers1