what does this generic syntax mean?

Question

public final class ImmutableList<E> { public final E head; public final ImmutableList<E> tail; public ImmutableList() { this.head = null; this.tail = null; } private ImmutableList(E head, ImmutableList<E> tail) { this.head = head; this.tail = tail; }

I know that public final E head is declaring an attribute from the generic tip E , this syntax is familiar to me but what does this public final ImmutableList<E> tail; mean, why declaring this attribute using the name of the generic class and what's the difference between :

public final E head;

and this :

public final ImmutableList<E> tail;

are they similar ?

why not using public final E tail istead of ImmutableList<E> Tail !! — Mouheb
– Mouheb, Commented Mar 2, 2015 at 19:52
That would not form a LinkedList, which is what this code is supposed to represent I would guess. — Rohit Jain
– Rohit Jain, Commented Mar 2, 2015 at 19:53
no the goal of the code to not use List and LinkedList to improve performance — Mouheb
– Mouheb, Commented Mar 2, 2015 at 19:54

wvdz · Accepted Answer · 2015-03-02 19:52:13Z

This code is a typical implementation of a recursive list. Each list has a head of type E, and as tail another recursive list which also has a head of type E, and as tail yet another recursive list.

The problem with recursion is that to understand it you have to understand recursion first.

René Winkler · Accepted Answer · 2015-03-02 19:52:17Z

The difference is that the former defines exactly one object of a type E named head. The latter defines an ImmutableList containing a bunch of objects E named tail. This is a recursive definition.

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