Is it possible to define a class with <T extends Something> and also use this class as paramter of a method as class<?>

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I saw a code in android that defines the following class:

android.widget.AdapterView<T extends android.widget.Adapter>

and also has the following interface: AdapterView.OnItemClickListener which contains the following method to be implemented:

onItemClick(AdapterView<?> parent, View view, int position, long id)

I can't understand how the first parameter works, if AdapterView was defined with constraint ( extneds...) How can it be a quiestion mark if it must be Adatper or something that derived from it?

android generics java
1 Answers

Class AdapterView<T extends android.widget.Adapter> requires that the concrete type used to create its instance extends Adapter and thus Adapter methods can be used inside.

Interface declaration (which actually doesn't have to be inside this class) accepts parameter AdapterView<?> where ? is so called wildcard. It means that type is unknown and can be anything.

For class it makes difference because its implementation will actually do something with type that is passed as generic parameter. You can define upper or lower bound on the type, but can't use unknown type in this case.

In case of interface - it is not a generic itself. Also, it's method is not generic, but accepts AdapterView of any type. Because of the wildcard any AdapterView can be passed to onItemLongClick. This is because AdapterView<A> is not a subclass of AdapterView<B>, even if A is a subclass of B. It is explained here (with example):

https://docs.oracle.com/javase/tutorial/java/generics/subtyping.html

You may want to read more about Java generics here: How do generics of generics work?

or/and here:

https://docs.oracle.com/javase/tutorial/java/generics/index.html

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