Iterable<String> iterable = asList(sourceArray);
    String[] newArray = Iterables.toArray(iterable, String.class);
    assertTrue(Arrays.equals(sourceArray, newArray));
  }

  public void testAny() {
    List<String> list = newArrayList();
    Predicate<String> predicate = Predicates.equalTo("pants");

    assertFalse(Iterables.any(list, predicate));
    list.add("cool");

      @Override
      protected Iterator<Integer> newTargetIterator() {
        return Iterators.filter(list.iterator(), isEven);
      }
    }.test();
  }

  public void testAny() {
    List<String> list = Lists.newArrayList();
    Predicate<String> predicate = Predicates.equalTo("pants");

    assertFalse(Iterators.any(list.iterator(), predicate));
    list.add("cool");

   *
   * However, for a variety of reasons, Google developers have written a ton of code over the past
   * decade that assumes that they can use checkNotNull for non-precondition checks. I had hoped to
   * take a principled stand on this, but the amount of such code is simply overwhelming. To avoid
   * creating a lot of compile errors that users would not find to be informative, we're giving in

   * provided key-value mapping, this method has no effect.
   *
   * <p>Note that a successful call to this method could cause the size of the bimap to increase by
   * one, stay the same, or even decrease by one.
   *
   * <p><b>Warning:</b> If an existing entry with this value is removed, the key for that entry is
   * discarded and not returned.
   *

many lessons to learn!  No, I've made up my mind about it; if I'm
Mabel, I'll stay down here!  It'll be no use their putting their
heads down and saying "Come up again, dear!"  I shall only look
up and say "Who am I then?  Tell me that first, and then, if I
like being that person, I'll come up:  if not, I'll stay down
here till I'm somebody else"--but, oh dear!' cried Alice, with a

  }

  /**
   * Returns an immutable list containing the given elements, sorted according to their natural
   * order. The sorting algorithm used is stable, so elements that compare as equal will stay in the
   * order in which they appear in the input.
   *
   * <p>If your data has no duplicates, or you wish to deduplicate elements, use {@code

        return;
      }

      /*
       * Reclassify nodes in each list to new Map. Because we are using power-of-two expansion, the
       * elements from each bin must either stay at same index, or move with a power of two offset.
       * We eliminate unnecessary node creation by catching cases where old nodes can be reused

f!.&a&d&amay?eos??g&no?ok?usak??hiku?k&awayim?uzii??ma&kan?y&asih?im???rawak?t&a&gon?ka&h?num?t???umo??wa&g&a&kan?nay?t??ias??ko!rih???y&ihsa?usak???e&m&ay?uruk??taruk?us??i&a&nohs?raihcat??goruk?h&cukuf?s&a&gih?hukuy??in???k&a&gako?muzim??iust?o?ustani??m&anim?otihsoynihs?u??r&ogo?ugasas??usu??ne&siek?zu&b?kihc???o&gukihc?h&ak?ot?ukihc??j&ono?ukihc??kayim?nihsukihc?to?uhc??u&fiazad?gnihs?stoyot????zihs!.&a&bmetog?d&amihs?eijuf?ihsoy?omihs??kouzihs?mihsim?ra&biah?honikam??tawi?wa&g&ekak?ukik??ki...

   * com.google.common.io.ByteStreams#newDataOutput()} to get a growable buffer.
   */
  public static byte[] toByteArray(long value) {
    // Note that this code needs to stay compatible with GWT, which has known
    // bugs when narrowing byte casts of long values occur.
    byte[] result = new byte[8];
    for (int i = 7; i >= 0; i--) {
      result[i] = (byte) (value & 0xffL);

many lessons to learn!  No, I've made up my mind about it; if I'm
Mabel, I'll stay down here!  It'll be no use their putting their
heads down and saying "Come up again, dear!"  I shall only look
up and say "Who am I then?  Tell me that first, and then, if I
like being that person, I'll come up:  if not, I'll stay down
here till I'm somebody else"--but, oh dear!' cried Alice, with a