What does int[]... arrays mean in Java?

Question

Can you help me? What does int[]... arrays mean in Java? Example:

public static int[] concat(int[]... arrays) {
int length = 0;
for (int[] array : arrays) {
    length += array.length;
}

See http://docs.oracle.com/javase/1.5.0/docs/guide/language/varargs.html — G.S, Dec 18 '13 at 09:08
Or here: http://mathbits.com/MathBits/Java/arrays/Declare.htm Sounds like you need to start right from the start. I googled that in seconds by the way. You should do the same. — Ciaran Gallagher, Dec 18 '13 at 09:09

score 4 · Answer 1 · edited Dec 18 '13 at 09:16

That is called varargs.. notation,

So that you can pass individual int[] objects to that method,, without worrying of no.of arguments.

When you write

public static int[] concat(int[]... arrays) {

Now you can use that method like

Classname.concat(array1,array2,array3)   //example1
Classname.concat(array1,array2)    //example2
Classname.concat(array1,array2,array3,array4)   //example3

A clear benefit is you need not to prepare an array to pass a method. You can pass values of that array directly.

score 1 · Answer 2 · edited May 23 '17 at 12:32

1

It means that the concat function can receive zero or more arrays of integers (int[]). That's why you can loop over the arrays argument, accessing one of the arrays contained in each iteration - if any. This is called variable arguments (or varargs).

Check this other question for more info.

edited May 23 '17 at 12:32

Community

1
1

answered Dec 18 '13 at 09:11

jnovo

5,659
2
38
56

score 1 · Answer 3 · answered Dec 18 '13 at 09:13

This means, that you can pass zero, one or more arrays of int (int[]) to your method. Consider following example

public void method(String ... strings)

can be called as

method()
method("s")
method("s", "d")
...

So your method can be called as

concat()
concat(new int[0])
concat(new int[0], new int[0])
...

Maroun · Answer 4 · 2013-12-18T09:18:54.503

1

This means that you can pass the method any number of int[] objects:

concat(new int[]{1,2});
concat(new int[]{1,2}, new int[]{3,4,5});
concat(new int[]{1}, new int[]{3,4,5,6}, new int[]{1,1,1,1,1,1,1});

Why this is useful? It's clear why :)

edited Dec 18 '13 at 09:18

answered Dec 18 '13 at 09:13

Maroun

94,125
30
188
241

score 0 · Answer 5 · answered Dec 18 '13 at 09:10

The ... is the most important part, it means you can have an unlimited amount of integers passed in and it is just accessed via the array, so you could call it as:

concat(1,2,3,4,5) or concat(1)

Most languages have this sort of feature, some call it var args others call it params.

score 0 · Answer 6 · answered Dec 18 '13 at 09:14

varargs ("...") notation basically informs that the arguments may be passed as a sequence of arguments (here - ints) or as an array of arguments. In your case, as your argument is of type "array of int", it means arguments may be passed as a sequence of arrays or as an array of arrays of int (note that array of arrays is quite equivalent to multidimensional array in Java).

score 0 · Answer 7 · answered Dec 18 '13 at 09:16

For example you have a method:

void functionName (Object objects...)
{
  Object object1=objects[0];
  Object object2=objects[1];
  Object object3=objects[2];
}

You can use your method as:

Object object1,object2,object3;

functionName(object1,object2,object3);

score 0 · Answer 8 · answered Dec 18 '13 at 09:16

Using Variable Arguments

The ellipsis (...) identifies a variable number of arguments, and is demonstrated in the following summation method.

public static int[] concat(int[]... arrays)

Use method

concat(val1, val2, val3);
concat(val1, val2, val3, val3);

What does int[]... arrays mean in Java?

8 Answers8

Related