How to convert a List to variable argument parameter java

Question

I have a method which takes a variable length string (String...) as parameter. I have a List<String> with me. How can I pass this to the method as argument?

[Varargs In Java: Variable Argument Method In Java](http://viralpatel.net/blogs/varargs-in-java-variable-argument-method-in-java-5/) — Grijesh Chauhan, May 25 '13 at 10:21
I am not sure if I understand your question correctly. Is it `method(String... length)` or `method(int length, String... param)`, or maybe something else? — Pshemo, May 25 '13 at 10:27

score 96 · Accepted Answer · edited Jan 27 '14 at 09:46

96

String... equals a String[] So just convert your list to a String[] and you should be fine.

edited Jan 27 '14 at 09:46

Svish

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answered May 25 '13 at 10:12

FloF

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Additional note to OP: if you ever have a method that takes Object..., this scheme won't work because the String[] will be understood as one Object. – Marko Topolnik May 25 '13 at 10:46
@MarkoTopolnik thanks. Any idea how to get around that? – Rishab178 Mar 06 '19 at 21:01

Alpesh Gediya · Answer 2 · 2013-05-25T10:41:11.750

42

String ... and String[] are identical If you convert your list to array.

using

Foo[] array = list.toArray(new Foo[list.size()]);

or

Foo[] array = new Foo[list.size()];
list.toArray(array);

then use that array as String ... argument to function.

edited May 25 '13 at 10:41

answered May 25 '13 at 10:13

Alpesh Gediya

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This also works: `list.toArray(new Foo[]{})`; useful if you don't want to store a list in additional variable. – liosedhel May 30 '14 at 06:43
What is this `Foo` thing ? I used a `String` instead as in `getUnsecuredPaths().toArray(new String[]{})` – Stephane Oct 03 '18 at 14:56
1

Obviously this is an old comment, but to anyone else reading this who has the same question; `Foo` is a pretty common term used in programming as a placeholder for a value that can change. It is useful when you're providing a generic example such as the one above, as `Foo` can be replaced by whichever object you're working with (in this case, a String). You may also see `Bar` used when there are multiple placeholders in an example. – Andy Shearer Feb 26 '21 at 12:19

Searene · Answer 3 · 2018-10-22T14:35:03.593

24

You can use stream in java 8.

String[] array = list.stream().toArray(String[]::new);

Then the array can be used in the ...args position.

edited Oct 22 '18 at 14:35

answered Sep 29 '17 at 06:46

Searene

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list.toArray(new String[]{}) suffices – J. Allen Jan 18 '18 at 17:14
FYI, if developing for Android streams require API 24 and jdk 1.8, so make sure its set up in your manifest and project settings. – Shane Sepac Apr 04 '18 at 23:52

score 1 · Answer 4 · edited Apr 01 '22 at 06:53

1

in case of long (primitive value)

long[] longArray = list.stream().mapToLong(o->g.getValue()).toArray();

when getValue() returns a long type

edited Apr 01 '22 at 06:53

Procrastinator

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answered Mar 31 '22 at 00:26

Samuel Morales

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score 1 · Answer 5 · answered May 19 '22 at 12:44

For Java > 8

You can very well use the toArray method very simply, as in:

String[] myParams = myList.toArray(String[]::new);

Streaming becomes really interesting when the elements of the List must be transformed, filtered or else before being converted to Array:

String[] myMappedAndFilteredParams = myList.stream()
                                 .map(param -> param + "some_added_stuff")
                                 .filter(param -> param.contains("some very important info"))
                                 .collect(toList())
                                 .toArray(String[]::new);

How to convert a List to variable argument parameter java

5 Answers5

Linked