}

    // Version 3 Referral Tests

    @Test
    public void testDecodeVersion3WithoutNameList() {
        // Prepare test data
        ByteBuffer bb = ByteBuffer.wrap(testBuffer).order(ByteOrder.LITTLE_ENDIAN);

        // Header
        bb.putShort((short) 3); // version
        bb.putShort((short) 100); // size (large enough for all strings)
        bb.putShort((short) 1); // serverType

   * they were inserted into the builder. For example, in the above example, {@code
   * WORD_TO_INT.entrySet()} is guaranteed to iterate over the entries in the order {@code "one"=1,
   * "two"=2, "three"=3}, and {@code keySet()} and {@code values()} respect the same order. If you
   * want a different order, consider using {@link #orderEntriesByValue(Comparator)}, which changes

     * {@link InvokerRequest#coreExtensions()} is in precedence order. Also reports conflicts, if any.
     * Finally, at DEBUG level reports configured vs selected extensions.
     */
    protected List<CoreExtension> selectCoreExtensions(C context, List<CoreExtensions> configuredCoreExtensions) {
        context.logger.debug("Configured core extensions (in precedence order):");
        for (CoreExtensions source : configuredCoreExtensions) {

      }

      @Override
      public List<String> order(List<String> insertionOrder) {
        List<String> order = new ArrayList<>();
        for (String s : insertionOrder) {
          int index = order.indexOf(s);
          if (index == -1) {
            order.add(s);
          } else {
            order.add(index, s);
          }
        }
        return order;
      }
    };
  }

  public void testCreate() {

    @Override
    public Iterable<Entry<E>> order(List<Entry<E>> insertionOrder) {
      // We mimic the order from gen.
      Map<E, Entry<E>> map = new LinkedHashMap<>();
      for (Entry<E> entry : insertionOrder) {
        map.put(entry.getElement(), entry);
      }

      Set<E> seen = new HashSet<>();
      List<Entry<E>> order = new ArrayList<>();
      for (E e : gen.order(new ArrayList<E>(map.keySet()))) {

    return new ImmutableLongArray(new long[] {e0});
  }

  /** Returns an immutable array containing the given values, in order. */
  public static ImmutableLongArray of(long e0, long e1) {
    return new ImmutableLongArray(new long[] {e0, e1});
  }

  /** Returns an immutable array containing the given values, in order. */
  public static ImmutableLongArray of(long e0, long e1, long e2) {

      }

      @Override
      public List<String> order(List<String> insertionOrder) {
        List<String> order = new ArrayList<>();
        for (String s : insertionOrder) {
          int index = order.indexOf(s);
          if (index == -1) {
            order.add(s);
          } else {
            order.add(index, s);
          }
        }
        return order;
      }
    };
  }

        scoreUpdater.addScoreBooster(booster2);
        scoreUpdater.addScoreBooster(booster3);

        // Clear execution order before test
        PriorityTrackingBooster.clearExecutionOrder();

        scoreUpdater.execute();

        // Boosters should be executed in priority order (3, 2, 1)
        List<Integer> executionOrder = PriorityTrackingBooster.getExecutionOrder();
        assertEquals(3, executionOrder.size());

      }

      @Override
      public List<String> order(List<String> insertionOrder) {
        List<String> order = new ArrayList<>();
        for (String s : insertionOrder) {
          int index = order.indexOf(s);
          if (index == -1) {
            order.add(s);
          } else {
            order.add(index, s);
          }
        }
        return order;
      }
    };
  }

   * dispatching an event, guaranteeing that all events posted on a single thread are dispatched to
   * all subscribers in the order they are posted.
   *
   * <p>When all subscribers are dispatched to using a <i>direct</i> executor (which dispatches on
   * the same thread that posts the event), this yields a breadth-first dispatch order on each
   * thread. That is, all subscribers to a single event A will be called before any subscribers to