How are Java generics different from C++ templates? Why can't I use int as a parameter?

Question

I am trying to create

ArrayList<int> myList = new ArrayList<int>(); 

in Java but that does not work.

Can someone explain why int as type parameter does not work?
Using Integer class for int primitive works, but can someone explain why int is not accepted?

Java version 1.6

Answer 1

Java generics are so different from C++ templates that I am not going to try to list the differences here. (See What are the differences between “generic” types in C++ and Java? for more details.)

In this particular case, the problem is that you cannot use primitives as generic type parameters (see JLS §4.5.1: "Type arguments may be either reference types or wildcards.").

However, due to autoboxing, you can do things like:

List<Integer> ints = new ArrayList<Integer>(); ints.add(3); // 3 is autoboxed into Integer.valueOf(3) 

So that removes some of the pain. It definitely hurts runtime efficiency, though.

Answer 2

The reason that int doesn't work, is that you cannot use primitive types as generic parameters in Java.

As to your actual question, how C++ templates are different from Java generics, the answer is that they're really, really different. The languages essentially apply completely different approaches to implementing a similar end effect.

Java tends to focus on the definition of the generic. That is, the validity of the generic definition is checked by only considering the code in the generic. If parameters are not properly constrained, certain actions cannot be performed on them. The actual type it's eventually invoked with, is not considered.

C++ is the opposite. Only minimal verification is done on the template itself. It really only needs to be parsable to be considered valid. The actual correctness of the definition is done at the place in which the template is used.