*       tasks -- even tasks that are not themselves {@code directExecutor} tasks.
   *   <li>If many such tasks are chained together (such as with {@code
   *       future.transform(...).transform(...).transform(...)....}), they may overflow the stack.
   *       (In simple cases, callers can avoid this by registering all tasks with the same {@link
   *       MoreExecutors#newSequentialExecutor} wrapper around {@code directExecutor()}. More

   * 1}, this method has the identical effect to {@link #add(Object)}. This method is functionally
   * equivalent (except in the case of overflow) to the call {@code
   * addAll(Collections.nCopies(element, occurrences))}, which would presumably perform much more
   * poorly.
   *

         */
        Runnable localTask = requireNonNull(task);
        task = null;
        localTask.run();
        return;
      }
      // Executor called reentrantly! Make sure that further calls don't overflow stack. Further
      // reentrant calls will see that their current thread is the same as the one set in
      // latestTaskQueue, and queue rather than calling execute() directly.

      setFutureSuccess.set(false);
      cancellationSuccess.set(false);
      finalResults.clear();
    }
    executor.shutdown();
  }

  // In a previous implementation this would cause a stack overflow after ~2000 futures chained
  // together.  Now it should only be limited by available memory (and time)
  public void testSetFuture_stackOverflow() {
    SettableFuture<String> orig = SettableFuture.create();

   * 1}, this method has the identical effect to {@link #add(Object)}. This method is functionally
   * equivalent (except in the case of overflow) to the call {@code
   * addAll(Collections.nCopies(element, occurrences))}, which would presumably perform much more
   * poorly.
   *

      /*
       * It speeds things up considerably to unconditionally add count to totalCount here,
       * but that destroys failure atomicity in the case of count overflow. =(
       */
      int cmp = comparator.compare(e, getElement());
      if (cmp < 0) {
        AvlNode<E> initLeft = left;
        if (initLeft == null) {
          result[0] = 0;

        void testLargeMidValue() {
            long largeMid = Long.MAX_VALUE;
            testMessage.setMid(largeMid);
            assertEquals(largeMid, testMessage.getMid());

            // HashCode should handle overflow correctly
            int hash = testMessage.hashCode();
            assertEquals((int) largeMid, hash);
        }

        @Test
        @DisplayName("Should handle negative status codes")

    return (byte) octet;
  }

  /** Returns a -1 if unable to parse */
  private static int tryParseDecimal(String string, int start, int end) {
    int decimal = 0;
    int max = Integer.MAX_VALUE / 10; // for int overflow detection
    for (int i = start; i < end; i++) {
      if (decimal > max) {
        return -1;
      }
      decimal *= 10;
      int digit = Character.digit(string.charAt(i), 10);
      if (digit < 0) {

      setFutureSuccess.set(false);
      cancellationSuccess.set(false);
      finalResults.clear();
    }
    executor.shutdown();
  }

  // In a previous implementation this would cause a stack overflow after ~2000 futures chained
  // together.  Now it should only be limited by available memory (and time)
  public void testSetFuture_stackOverflow() {
    SettableFuture<String> orig = SettableFuture.create();

    i = 1 << 30;
    assertFalse(MinMaxPriorityQueue.isEvenLevel(i - 2));
    assertTrue(MinMaxPriorityQueue.isEvenLevel(i - 1));
    assertTrue(MinMaxPriorityQueue.isEvenLevel(i));

    // 1 << 31 is negative because of overflow, 1 << 31 - 1 is positive
    // since isEvenLevel adds 1, we need to do - 2.
    assertTrue(MinMaxPriorityQueue.isEvenLevel((1 << 31) - 2));
    assertTrue(MinMaxPriorityQueue.isEvenLevel(Integer.MAX_VALUE - 1));