Dtjytyk

I've recently come across this unusual (to me) Java syntax...here's an example of it:

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

Notice the positioning of the <String, Long> type arguments...it's not after the type as normal but before. I don't mind admitting I've never seen this syntax before. Also note there are 2 type arguments when ArrayList only has 1.

Does the positioning of the type arguments have the same meaning as putting them after the type? If not, what does the different positioning mean?

Why is it legal to have 2 type arguments when ArrayList only has 1?

I've searched the usual places, eg. Angelika Langer and on here but can't find any mention of this syntax anywhere apart from the grammar rules in the Java grammar file on the ANTLR project.

This is unusual alright, but fully valid Java. A class may have a generic constructor, for example:

public class TypeWithGenericConstructor 

 public <T> TypeWithGenericConstructor(T arg) 
 // TODO Auto-generated constructor stub
 

I suppose that more often than not we don’t need to make the type argument explicit. For example:

 new TypeWithGenericConstructor(LocalDate.now(ZoneId.systemDefault()));

Now T is clearly LocalDate. However there may be cases where Java cannot infer (deduce) the type argument. Then we supply it explicitly using the syntax from your question:

 new <LocalDate>TypeWithGenericConstructor(null);

Of course we may also supply it even though it is not necessary if we think it helps readability or for whatever reason:

 new <LocalDate>TypeWithGenericConstructor(LocalDate.now(ZoneId.systemDefault()));

In your question you seem to be calling the java.util.ArrayList constructor. That constructor is not generic (only the ArrayList class as a whole is, that’s something else). Why Java allows you to supply type arguments in the call when they are not used, I don’t know, but it does. My Eclipse gives me a warning:

Unused type arguments for the non generic constructor ArrayList() of
type ArrayList; it should not be parameterized with arguments

But it’s not an error, and the program runs fine (I additionally get warnings about missing type arguments for List and ArrayList, but that again is a different story).

Does the positioning of the type arguments have the same meaning as
putting them after the type? If not, what does the different
positioning mean?

No, it’s different. The usual type argument/s after the type (ArrayList<Integer>) are for the generic class. The type arguments before are for the * constructor*.

The two forms may also be combined:

 List<Integer> list = new <String, Long>ArrayList<Integer>();

I would consider this a bit more correct since we can now see that the list stores Integer objects (I’d still prefer to leave out the meaningless <String, Long>, of course).

Why is it legal to have 2 type arguments when ArrayList only has 1?

First, if you supply type arguments before the type, you should supply the correct number for the constructor, not for the class, so it hasn’t got anything to do with how many type arguments the ArrayList class has got. That really means that in this case you shouldn’t supply any since the constructor doesn’t take type arguments (it’s not generic). When you supply some anyway, they are ignored, which is why it doesn’t matter how many or how few you supply (I repeat, I don’t know why Java allows you to supply them meaninglessly).

Here:

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

You have a raw type, as you don't specify a generic type for your list object on the left hand side. Providing the generic type on the right hand side is mostly not required in the first place, as the compiler can defer the generic type.

A bit of guessing here: due to fact that list itself is a raw type, the type specification given to new is somehow ignored. The other answer nicely explains why you can have type parameters in that place. But because a raw type doesn't care about type parameters, the compiler doesn't bother checking that detail here.