if (exception == null) {
        exception = ioe
      }
    }
  }
  if (exception != null) {
    throw exception
  }
}

internal fun <E> MutableList<E>.addIfAbsent(element: E) {
  if (!contains(element)) add(element)
}

internal fun Exception.withSuppressed(suppressed: List<Exception>): Throwable =
  apply {
    for (e in suppressed) addSuppressed(e)
  }

   * <i>should</i> hold {@code expectedSize} elements without growth.
   *
   * @param expectedSize the number of elements you expect to add to the returned set
   * @return a new, empty {@code ObjectCountHashMap} with enough capacity to hold {@code
   *     expectedSize} elements without resizing
   * @throws IllegalArgumentException if {@code expectedSize} is negative
   */

  }

  public static class Chars extends SampleElements<Character> {
    public Chars() {
      // elements aren't sorted, to better test SortedSet iteration ordering
      super('b', 'a', 'c', 'd', 'e');
    }
  }

  public static class Enums extends SampleElements<AnEnum> {
    public Enums() {
      // elements aren't sorted, to better test SortedSet iteration ordering
      super(AnEnum.B, AnEnum.A, AnEnum.C, AnEnum.D, AnEnum.E);

        multimap.putAll("foo", asList(7, 8));
        return multimap.get("foo").iterator();
      }

      @Override
      protected void verify(List<Integer> elements) {
        assertEquals(newHashSet(elements), multimap.get("foo"));
      }
    }.test();
  }

  @GwtIncompatible // unreasonably slow
  public void testEntriesIteration() {
    Set<Entry<String, Integer>> set =

    //     ends up at a[d], which in turn ends up at a[2d], and so on until we get back to a[0].
    //     (All indices taken mod n.) If d and n are mutually prime, all elements will have been
    //     moved at that point. Otherwise, we can rotate the cycle a[1], a[1 + d], a[1 + 2d], etc,
    //     then a[2] etc, and so on until we have rotated all elements. There are gcd(d, n) cycles

      numCallsToNewTargetIterator++;
      return Lists.newArrayList(1).iterator();
    }

    @Override
    protected void verify(List<Integer> elements) {
      numCallsToVerify++;
      super.verify(elements);
    }
  }

  public void testVerifyGetsCalled() {
    TesterThatCountsCalls tester = new TesterThatCountsCalls();

    tester.test();

    assertEquals(

  public void testSet_replacingNull() {
    E[] elements = createSamplesArray();
    int i = aValidIndex();
    elements[i] = null;
    collection = getSubjectGenerator().create(elements);

    doTestSet(e3());
  }

  private void doTestSet(E newValue) {
    int index = aValidIndex();
    E initialValue = getList().get(index);
    assertEquals(
        "set(i, x) should return the old element at position i.",
        initialValue,

                  @Override
                  protected NavigableSet<String> create(String[] elements) {
                    TreeMultimap<String, Integer> multimap =
                        TreeMultimap.create(Ordering.natural().nullsFirst(), Ordering.natural());
                    for (int i = 0; i < elements.length; i++) {
                      multimap.put(elements[i], i);
                    }
                    return multimap.keySet();

 * <i>first</i> and <i>last</i> are considered to refer to the elements' ordering in a list.
 *
 * @author Louis Wasserman
 */
@GwtCompatible
@ElementTypesAreNonnullByDefault
final class SortedLists {
  private SortedLists() {}

  /**
   * A specification for which index to return if the list contains at least one element that
   * compares as equal to the key.
   */

  }

  /**
   * Reverses the elements of {@code array}. This is equivalent to {@code
   * Collections.reverse(Booleans.asList(array))}, but is likely to be more efficient.
   *
   * @since 23.1
   */
  public static void reverse(boolean[] array) {
    checkNotNull(array);
    reverse(array, 0, array.length);
  }

  /**