Java coding conventions - what are these curly braces for?

Question

I'm new to JAVA and I've run into the following code snippet.

My question is about the curly braces that surrounds the 2 setter calls.

I can't find anything in the coding conventions http://www.oracle.com/technetwork/java/codeconventions-150003.pdf that explains this code snippet.

Other than to scope the setter calls (but I can't understand the need for that), I can't think of any reason for the curly braces, can someone please explain the purpose of the curly braces? (sorry I didn't have any luck looking for previous questions on this)

Foobar foobar = FoobarManager.getFoobar();
{
    foobar.setName("MyFoo");
    foobar.setTimestamp(timestamp);
}

Is this code inside a method or constructor, or inside only a class? — rgettman, Jun 23 '15 at 18:11
Inside a method, in fact I see this style in a number of places in this project I'm looking at, all of them are inside methods. — liaaba, Jun 23 '15 at 18:12
It creates a separate scope (so the same variable name can be used twice or just for safety). It's very rarely used because it's a good sign you need to split your method up. The way it's used there though it is entirely pointless — Richard Tingle, Jun 23 '15 at 18:14
They create a code *block*, which has limited value (you can scope local variables to them), but in this example you can remove the braces with absolutely no consequences. — Bohemian, Jun 23 '15 at 18:15
See this question: https://stackoverflow.com/questions/1563030/anonymous-code-blocks-in-java — Jonathan Crosmer, Jun 23 '15 at 18:27
Check this question too http://stackoverflow.com/questions/4514583/is-there-a-use-of-free-floating-block-inside-a-method-in-java — Alejandro Agapito Bautista, Jun 23 '15 at 18:45

Alejandro Agapito Bautista · Answer 1 · 2015-06-23T18:29:48.367

1

This is an java initializer block and it is going to work after the super call of your constructor, you can execute the next code

public class Test {
    public Test() {
        super();
        System.out.println("Hello");
    }

    {
        System.out.println("Bye");
    }

    public static void main(String[] args) {
        new Test();
    }
}

If you check the output, the result is :

Bye
Hello

This is because fist is going to be executed the super call, after this is going to be executed the initializer block and after this the next lines of your constructor.

It is helpful if you have many constructors and you want to execute some lines of code independently of which constructor was executed

edited Jun 23 '15 at 18:29

answered Jun 23 '15 at 18:19

Alejandro Agapito Bautista

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What you're showing there is an _initializer block_ and not an _anonymous block_. – Tom Jun 23 '15 at 18:25
Thanks for the reply. I understand the initializer block concept, but it is not clear to me how the snippet I posted work as an anonymous block, it resides inside a method and not the body of the class. I might be missing something... – liaaba Jun 23 '15 at 18:26

Java coding conventions - what are these curly braces for?

1 Answers1