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Just came across this statement and was wondering why this function call had what at first looked like a cast?

SomeClass bo = new SomeClass(); // blabla something like that to initialize the object variable
(bo).setValue(bo.getValue().negate());

As I have not yet seen this syntax - what does it do compared to a simple

bo.setValue(bo.getValue().negate());

?

Lino
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PhiSe
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  • This looks like an anonymous class. But I fail to see the point in an anonymous subclass of `Object` . – Arnaud Jul 27 '18 at 07:31
  • the braces only enforce precedence of an expression over another operator with usually higher precedence, eg. `(1 + 2) * 3`. Here they do not change the meaning since there is no operator in the expression inside it* – joH1 Jul 27 '18 at 07:33
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    @Arnaud I think the issue is on the second line... The first is just an example of initialization--misleading though. – joH1 Jul 27 '18 at 07:33
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    There is no difference between `(bo).setValue(...)` and `bo.setValue(...)`. The parenthesis around `bo` are superfluous. – Andreas Jul 27 '18 at 07:35
  • Question title is misleading. Your question is about the `()` around `bo`. `()` are called "parentheses" (or "round brackets"), while `{}` are called "braces" (or "curly brackets"). See [Bracket vs brace](https://english.stackexchange.com/questions/3379/bracket-vs-brace). – Andreas Jul 27 '18 at 07:41

1 Answers1

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(bo).setValue(bo.getValue().negate()) and bo.setValue(bo.getValue().negate()) are identical statements and the parentheses are reduntant here.

They are needed though when we write expressions like 

Object o;
(o = new Object()).toString();  // class java.lang.Object

If we had omitted them, 

Object o;
o = new Object().toString();  // class java.lang.String
Andrew Tobilko
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