String... parameter in Java

Question

I have to implement an API for a homework assignment, and my instructor has used a notation I am unfamiliar with for one of the methods in the API (javadoc based).

public void method(String... strs);

What do the '...' mean? It later looks like I'll need to call this same method using a single string actual parameter, as well as multiple string actual parameters...

Java doesn't have optional arguments (to my knowledge), so I am a little confused here...

dagge · Accepted Answer · 2012-01-28T22:27:58.517

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It's called varargs; http://docs.oracle.com/javase/6/docs/technotes/guides/language/varargs.html

It means you can pass an arbitrary number of arguments to the method (even zero).

In the method, the arguments will automatically be put in an array of the specified type, that you use to access the individual arguments.

edited Jan 28 '12 at 22:27

answered Jan 28 '12 at 21:23

dagge

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Ah, thanks for the quick response. I hadn't seen this before... Very nice. – Desh Banks Jan 28 '12 at 21:59
I would just add that this type of function is called **variadic**: http://en.wikipedia.org/wiki/Variadic_function – amphibient Oct 10 '13 at 20:36

score 6 · Answer 2 · answered Jan 28 '12 at 21:24

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It's called an ellipsis and it means the method can take multple Strings as its argument.

See: The Java tutorial on passing arguments on Oracle's site.

answered Jan 28 '12 at 21:24

Brian Roach

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score 5 · Answer 3 · edited Aug 28 '20 at 07:52

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Yes, that means you can take arbitrary no of Strings as an argument for this method.

For your method:

public void method(String... strs);

You can call it as:

method(str)
method(str1, str2)
method(str1,str2,str3)

Any no of arguments would work. In other words, it is a replacement for:

 public void method(String[] str);

edited Aug 28 '20 at 07:52

mehmet ali akbay

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answered Jan 28 '12 at 21:25

Johnydep

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Uh, no, it doesn't (you're half correct - your first sentence is the problem). – Brian Roach Jan 28 '12 at 21:28
@BrianRoach, Oh yes sorry you are right :) – Johnydep Jan 28 '12 at 21:31
Haven't found a max limit of how many arguments varargs can take, but it's definitely more than three – Kennet Jan 28 '12 at 21:33
+1 for coming back and fixing it. – Brian Roach Jan 28 '12 at 22:22
Thanks for the example. Upon knowing about 'varargs' I've also found that the vararg argument must be the final formal parameter declared in the parameter list of the function using varargs. – Desh Banks Feb 05 '12 at 17:43

score 1 · Answer 4 · edited May 23 '17 at 12:32

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See java optional parameters : as of Java 5, Java has support for variable numbers of arguments.

edited May 23 '17 at 12:32

Community

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answered Jan 28 '12 at 21:24

Borealid

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String... parameter in Java

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