Is there a way to open and close a stream easily at kotlin?

Question

What I have to do at java:

try(InputStream inputStream = new FileInputStream("/home/user/123.txt")) {

    byte[] bytes = new byte[inputStream.available()];
    inputStream.read(bytes);
    System.out.println(new String(bytes));


} catch (IOException e) {
    e.printStackTrace();
}

But kotlin doesn't know about try-with-resources! So my code is

try {
    val input = FileInputStream("/home/user/123.txt")
} finally {
    // but finally scope doesn't see the scope of try!
}

Is there an easy way to close the stream ? And I don't speak only about files. Is there a way to close any stream easily ?

score 37 · Accepted Answer · answered Sep 07 '17 at 13:52

37

Closeable.use is what you're looking for:

val result = FileInputStream("/home/user/123.txt").use { input ->
    //Transform input to X
}

answered Sep 07 '17 at 13:52

Kiskae

24,655
2
77
74

1

Does it also flush the stream? – Onkar Nene Nov 08 '19 at 10:38
Of course not. Just check the source code. – Yoann CAPLAIN Apr 02 '20 at 04:25

score -4 · Answer 2 · answered Sep 07 '17 at 14:38

-4

 BufferedInputStream input = null;
        try {
            input = new BufferedInputStream(new FileInputStream("/home/user/123.txt"));
        } catch (Exception e) {
            e.printStackTrace();
        } finally {
            if (input != null) {
                input.close();
            }
        }

Works fine with any Inputstream. This is just an example i used.

answered Sep 07 '17 at 14:38

Didi

29
5

5

This is java code – Erythrozyt Jan 28 '21 at 06:14

Is there a way to open and close a stream easily at kotlin?

2 Answers2