Java generics understanding

Question

Could you please explain to me what happens here and why it works without errors.

List list = new ArrayList<String>();
list.add(1);

I'm sorry, if there already were similar questions, I failed to properly formulate the question so can't google it

Do you have compiler warnings turned on? – shoover Dec 24 '13 at 19:19 — shoover, Dec 24 '13 at 19:19

score 4 · Accepted Answer · answered Dec 24 '13 at 19:25

4

What happens here exposes the weakness of type erasure, the mechanism behind Java generics. The implementation of ArrayList does not change when you change its generic type parameter: it's always java.lang.Object behind the scene.

Since you did not give a type parameter to the interface when declaring your list, compiler allows you to add objects of any type, not only Strings. In particular, int gets autoboxed into java.lang.Integer, so the value of 1 can be added to your list.

answered Dec 24 '13 at 19:25

Sergey Kalinichenko

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So, nothing at all depends on ArrayList type parameter? Only List's parameter matters? – Patison Dec 24 '13 at 19:47
@MaxVasileusky Correct, since the type gets erased after the compile time, and because the type of the variable does not depend on what you assign it, only the generic parameter of the List interface in the declaration matters. – Sergey Kalinichenko Dec 24 '13 at 20:06

Reimeus · Answer 2 · 2013-12-24T20:33:54.707

2

Using the raw type List list is approximately equivalent to using List<Object> without type checking. This type of List will accept any type of object. The statement

list.add(1);

autoboxes an Integer to add to the List

edited Dec 24 '13 at 20:33

answered Dec 24 '13 at 19:22

Reimeus

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That's due to generic invariance, have a look at this [snippet](http://ideone.com/C1AkJ7) – Reimeus Dec 24 '13 at 19:59
A List of unknowns (`?`) does not permit any addition (except for `null`)... – Reimeus Dec 24 '13 at 20:07
Actually List l is not even equivalent to List >. I messed up with the concept. Apology. – Aman Arora Dec 24 '13 at 20:09
Yes you are right. But it ain't equivalent to writing List – Aman Arora Dec 24 '13 at 20:10
Well it's not _exactly_ equivalent (as no warning is produced) but _is_ functionally equivalent – Reimeus Dec 24 '13 at 20:14

score 1 · Answer 3 · answered Dec 24 '13 at 19:20

Generics of the reference is important. So

List list = new ArrayList<String>();

makes no sense. It must be

List<String> list = new ArrayList<String>();

If you don not provide at generics for the reference type(not the actual object type) it is taken to be as Object. So you can add anything to it.

Also generics is a compile time phenomenon. So when you say how it works it means how your code is compiling without compilation errors. It is for a simple reason that no generics associated with the reference Object and hence is taken as Object(default).

score 1 · Answer 4 · answered Dec 24 '13 at 19:23

Could you please explain me what happens here and why it works without errors.

Type erasure.

Java generics exist only at compile time. If you try to shove an int into something that is only declared as List, (which is bound to Objects by default) at runtime, it will work just fine.

score 0 · Answer 5 · answered Dec 24 '13 at 19:23

Sure. if you're familiar with C++, generics are kinda like templates (with few exception, but lets not dig that detail). The main idea of Generics is for type safety in your code and hence clarity. This is great for static typed languages like Java. List is an interface and Arraylist extends from it. Hence its a parent (super class) is List. List has features like sublist, etc.

list.add(1) simply calls the add method of the list interface (which is overridden in Arraylist) to add that element to the array.

Also use:

List<String> list = new ArrayList<String>

great article on generics here: http://docs.oracle.com/javase/tutorial/java/generics/

Java generics understanding

5 Answers5