Constructor vs Curly Brackets (?)

Question

While searching for a way to define static constructors, I've stumbled upon the use of the following:

class MyClass {
    { /* code for constructor-like (?) effect */ }
}

Basically the code that is entered in curly brackets directly in the class works, as far as I can see, exactly the same as a code in a constructor - except that it is called before the constructor.

What is the purpose of this, and are there other differences between this and a standard constructor?

Thanks in advance,

Don't get confused with the curly brackets. Those are not necessary. — Sarwar Erfan, Nov 28 '11 at 12:13
@Sarwar Erfan: Nope; they are, otherwise you can't get "code with constructor like effect". For e.g. catching exceptions etc. — Sanjay T. Sharma, Nov 28 '11 at 12:15
@SanjayT.Sharma: Ok, they are called before constructor, so I would not call it "constructor like" effect. If you omit the curly brackets, you cannot catch excpetions, but aren't the calling time and result same? — Sarwar Erfan, Nov 28 '11 at 12:19

score 8 · Answer 1 · edited May 23 '17 at 10:34

8

Those are called instance initializers. The details can be found in another SO question.

edited May 23 '17 at 10:34

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answered Nov 28 '11 at 12:12

Sanjay T. Sharma

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Thanks. (Exactly same answer as Jesper, but he answered first so I have to give him the accepted answer) – Acidic Nov 28 '11 at 12:49

score 3 · Accepted Answer · edited May 23 '17 at 12:34

3

The block with the curly brackets is an instance initializer block. This page from Oracle's Java Tutorials has some more info about it.

Also see: How is an instance initializer different from a constructor?

edited May 23 '17 at 12:34

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answered Nov 28 '11 at 12:12

Jesper

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Constructor vs Curly Brackets (?)

2 Answers2