"new Foo(){}" vs. "new Foo()" in Java

Question

For example:

Object o1 = new ArrayList<String>();
Object o2 = new ArrayList<String>(){};
Object o3 = new ArrayList<String>(){{}};

What's the difference?

I can't google out the 2nd/3rd grammar of Java, any reference?

possible duplicate of [Interview : Can we instantiate abstract class?](http://stackoverflow.com/questions/13670991/interview-can-we-instantiate-abstract-class) — Rohit Jain, Sep 02 '13 at 18:08
Third one is declaring an instance initializer block of the anonymous class. — Sotirios Delimanolis, Sep 02 '13 at 18:11
Re: the third one, http://www.c2.com/cgi/wiki?DoubleBraceInitialization — Matt Ball, Sep 02 '13 at 18:12
Very interesting to see that third one, what do we call the third notation? — SimonT, Sep 02 '13 at 18:15
@SimonT Double-brace initialization, but try to stay away from it. — arshajii, Sep 02 '13 at 18:16

Peter Lawrey · Accepted Answer · 2013-09-02T18:56:54.623

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The first creates an ArrayList

The second creates an anonymous subclass of ArrayList which has a specific generic type of String

The third is the same but it has a initializer block which is empty.

Note: Where ever possible, you should write the simplest and clearest code you can, esp if you are thinking about performance.

edited Sep 02 '13 at 18:56

answered Sep 02 '13 at 18:11

Peter Lawrey

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I think you should add "unless there is a detailed comment explaining that there is some weird special need for an anonymous class to circumvent reflection, the 2nd and 3rd cases are just plain stupid, confusing, and wrong." – user949300 Sep 02 '13 at 18:23
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I'll add that [JMock](http://www.jmock.org) uses the third style for it's mocks to subclass and encode expectations. – Sled Sep 02 '13 at 19:33

nachokk · Answer 2 · 2013-09-10T15:31:30.763

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Object o1 = new ArrayList<String>();

Creates an ArrayList.

Object o2 = new ArrayList<String>(){};

Here you are creating an anonymous class that extends ArrayList<String> and don't override anything. So the difference it's that you are subclassing an ArrayList without overrding behaviour, never do this if you don't have a good reason.

Object o3 = new ArrayList<String>(){{}};

You are creating the same as 2 but with an empty initalizer block.

edited Sep 10 '13 at 15:31

answered Sep 02 '13 at 18:09

nachokk

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thanks! got it. maybe the second is useful to avoid "type erasure"? – marstone Sep 02 '13 at 18:14
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@marstone. No that's not the use. Type erasure cannot be avoided. – Rohit Jain Sep 02 '13 at 18:15
@RohitJain if not, how could google gson keep the generic parameter type by using "new TypeToken>() {}.getType()"? sorry if this is off the topic. – marstone Sep 02 '13 at 18:22
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@marstone they use black magic of reflection, it's unrelated to anonymous subclasses, you can find some details [here](https://code.google.com/p/guava-libraries/wiki/ReflectionExplained) – Katona Sep 02 '13 at 18:28
@Katona: It is *absolutely* related to anonymous subclasses; I'm not sure how you can claim otherwise. The subclassing is what allows the type-argument to be available at runtime despite erasure. (Sure, it's only available using the black magic of reflection, but without the subclassing even that much would not be possible.) – ruakh Sep 02 '13 at 23:15
@ruakh, ok, let me correct myself, it is related to _subclassing_ but doesn't have to be anonymous, and certainly doesn't prevent type erasure. – Katona Sep 03 '13 at 06:35

score 2 · Answer 3 · answered Sep 02 '13 at 18:13

Object o1 = new ArrayList<String>();

Creating a new ArrayList object and assigning it to o1

Object o2 = new ArrayList<String>(){};

Creating a new instance of an anonymous class that extends ArrayList and assigning it to o2

Object o3 = new ArrayList<String>(){{}};

Creating a new instance of a (different from o2) anonymous class that extends ArrayList that has a no-op instance initializer.

Functionally the anon classes assigned to o2 and o3 are equivalent but technically they will be different classes.

score 0 · Answer 4 · answered Sep 02 '13 at 18:11

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You are anonymously instantiating a class. You could override methods in the later case without the need to define a new class in an own file.

answered Sep 02 '13 at 18:11

sorencito

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score 0 · Answer 5 · answered Sep 02 '13 at 18:11

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o3 is still an instance of an anonymous subclass of ArrayList which has an empty instance initialization block (the inner {}).

answered Sep 02 '13 at 18:11

Katona

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score 0 · Answer 6 · answered Sep 02 '13 at 18:12

// Instantiate an object of type `ArrayList<String>`
Object o1 = new ArrayList<String>();

// Instantiate an anonymous subclass of `ArrayList<String>`
Object o2 = new ArrayList<String>(){};

// Instantiate an anonymous subclass of `ArrayList<String>` with an empty initializer block
Object o3 = new ArrayList<String>(){{}};

score 0 · Answer 7 · answered Sep 02 '13 at 18:12

Object o2 = new ArrayList<String>(){};
it is annonymous inner class that extends ArrayList class and overrides nothing.
Object o2 = new ArrayList<String>(){{}};
it is also annonymous inner class that extends ArrayList class and overrides nothing but has a empty instance block.

"new Foo(){}" vs. "new Foo()" in Java

7 Answers7