}

      @Override
      @ParametricNullness
      public E next() {
        final Map.Entry<E, Count> mapEntry = backingEntries.next();
        toRemove = mapEntry;
        return mapEntry.getKey();
      }

      @Override
      public void remove() {
        checkState(toRemove != null, "no calls to next() since the last call to remove()");
        size -= toRemove.getValue().getAndSet(0);

    checkNotNull(predicate);
    Iterator<T> iterator = removeFrom.iterator();
    while (iterator.hasNext()) {
      T next = iterator.next();
      if (predicate.apply(next)) {
        iterator.remove();
        return next;
      }
    }
    return null;
  }

  /**
   * Determines whether two iterables contain equal elements in the same order. More specifically,

                return Integer.valueOf(from);
              }
            });

    Iterator<Integer> resultIterator = result.iterator();
    resultIterator.next();

    try {
      resultIterator.next();
      fail("Expected NFE");
    } catch (NumberFormatException expected) {
    }
  }

  public void testNullFriendlyTransform() {

    int i = 0;
    while (expectedIterator.hasNext()) {
      Object expected = expectedIterator.next();

      assertTrue(
          "index " + i + " expected <" + expected + "., actual is exhausted",
          actualIterator.hasNext());

      Object actual = actualIterator.next();
      assertEquals("index " + i, expected, actual);
      i++;
    }
    if (actualIterator.hasNext()) {

 * acquire(1000)} will result in exactly the same throttling, if any), but it affects the throttling
 * of the <i>next</i> request. I.e., if an expensive task arrives at an idle RateLimiter, it will be
 * granted immediately, but it is the <i>next</i> request that will experience extra throttling,
 * thus paying for the cost of the expensive task.
 *
 * @author Dimitris Andreou
 * @since 13.0
 */

        AvlNode<E> next = current.succ();

        current.elemCount = 0;
        // Also clear these fields so that one deleted Entry doesn't retain all elements.
        current.left = null;
        current.right = null;
        current.pred = null;
        current.succ = null;

        current = next;
      }
      successor(header, header);

   * high 32 bits of each long is the "prev" pointer, whereas the low 32 bits is the "succ" pointer
   * (pointing to the next entry in the linked list). The pointers in [size(), entries.length) are
   * all "null" (UNSET).
   *
   * <p>A node with "prev" pointer equal to {@code ENDPOINT} is the first node in the linked list,
   * and a node with "next" pointer equal to {@code ENDPOINT} is the last node.
   */

  private int position;

  /** Returns the element with the specified index. This method is called by {@link #next()}. */
  @ParametricNullness
  protected abstract E get(int index);

  /**
   * Constructs an iterator across a sequence of the given size whose initial position is 0. That
   * is, the first call to {@link #next()} will return the first element (or throw {@link
   * NoSuchElementException} if {@code size} is zero).
   *

    assertEquals(o1, three.next().target);
    assertFalse(three.hasNext());

    three = registry.getSubscribers("");

    registry.unregister(s2);

    assertEquals(s1, three.next().target);
    assertEquals(s2, three.next().target);
    assertEquals(o1, three.next().target);
    assertFalse(three.hasNext());

    Iterator<Subscriber> two = registry.getSubscribers("");
    assertEquals(s1, two.next().target);

    assertEquals((Integer) 12, it.next());
    assertEquals((Integer) 30, it.next());
    assertEquals((Integer) 40, it.next());
    // Skipping 20
    assertEquals((Integer) 11, it.next());
    // Not skipping 400, because it moved back down
    assertEquals((Integer) 400, it.next());
    assertEquals((Integer) 13, it.next());
    assertEquals((Integer) 200, it.next());
    assertEquals((Integer) 300, it.next());