Re: Different treatment of the NodeType of LinkedList in C++ and Java

James Kanze <>
Thu, 29 Oct 2009 03:18:37 -0700 (PDT)
On Oct 28, 9:14 pm, "" <> wrote:

I am a student who is taking the data structure course. When
we are learning the data structure LinkedList, I noticed that
Java and C++ treated the type of the node of Linkedlist

In STL, C++ actually publishes the typed of list_node:
class List{
typedef __list_node<T> list_node:

I suppose you miscopied something, since the name of the class
in the standard is list, not List. And the "exposition" of the
node type is probably due to some techical implementation issue;
it's not officially exposed, as part of the documented interface
in the standard.

But in Java, the class of LinkedListNode has package access,
so the user of LinkedList won't even see this node type.

Package access is less protective than protected access in C++
(or in Java). With protected access, the only way to access the
name is to derive from the class. With package access, you can
either derive from the class, or declare another class as a
member of the same package.

Personally, I wound think the type of node is the
implementation detail, and the best practice is to hide it
from the user.

Agreed. On the other hand, there can be technical reasons for
exposing it, partially or completely: in my pre-standard
DLListOf, it was fully private, but in my pre-standard
AssocArrayOf, the node base type was a protected member of the
(non-templated) base class.

But I am posting this message to see, is there any particular
reason that C++ choose to publish the list_node type?

C++ doesn't publish it, or even require it to exist (although I
don't see how you could implement std::list without it).

Does this has something to do with the difference in this two

I don't really think so. I don't really see why it isn't
private in both cases, but I suspect that there are some
technical issues involved. (In the case of Java, it might be
protected or package in order to facilitate the implementation
of some more strongly typed derived class. In the case of C++,
I don't see any real reason.)

James Kanze

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