Re: Creating a byte[] of long size

Tom Anderson <>
Mon, 12 Jul 2010 12:52:41 +0100
On Sun, 11 Jul 2010, Patricia Shanahan wrote:

Tom Anderson wrote:

On Fri, 9 Jul 2010, Patricia Shanahan wrote:

On 7/9/2010 12:45 PM, Tom Anderson wrote:

On Thu, 8 Jul 2010, Eric Sosman wrote:

Or, you could have BigList implement List but "lie" in its .size()
method, in somewhat the same way TreeSet "lies" about the Set contract.

How does TreeSet lie about the Set contract?

The case I'm aware of involves a TreeSet with a Comparator, that is not
consistent with the .equals methods of the TreeSet elements. The TreeSet
always goes by the Comparator results. That means the TreeSet could
contain elements a and b such that a.equals(b).

Though that feel more like "TreeSet requires its Comparator to be
consistent with equals" than "TreeSet lies about the Set contract".

If i write [...] Is HashSet now breaking the Set contract?

But TreeSet does not do completely wild, uncontracted things given an
inconsistency, the way HashSet does given a hash that is inconsistent
with equals. TreeSet is perfectly usable given a Comparator that is
internally consistent, even if it is not consistent with equals.

Another way of looking at this is "TreeSet acts as though its Comparator
is consistent with equals.".

Well *another* way (actually, the first way again) of looking at it is
that TreeSet expects its Comparator to be consistent with equals, but,
like a world-weary Latin master, tolerates failure to meet its

Or perhaps a more neutral, and complete, way to put it is:

  Note that the ordering maintained by a set (whether or not an explicit
  comparator is provided) must be consistent with equals if it is to
  correctly implement the Set interface. (See Comparable or Comparator for
  a precise definition of consistent with equals.) This is so because the
  Set interface is defined in terms of the equals operation, but a TreeSet
  instance performs all element comparisons using its compareTo (or
  compare) method, so two elements that are deemed equal by this method
  are, from the standpoint of the set, equal. The behavior of a set is
  well-defined even if its ordering is inconsistent with equals; it just
  fails to obey the general contract of the Set interface.


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