// if waiting works, this should get stuck
      Queues.drain(q, newArrayList(), 1, MAX_VALUE, NANOSECONDS);
      fail();
    } catch (InterruptedException expected) {
      // we indeed waited; a slow thread had enough time to interrupt us
    }
  }

  // same as above; uninterruptible version
  @SuppressWarnings("ThreadPriorityCheck") // TODO: b/175898629 - Consider onSpinWait.

    /**
     * Careful here: if threads are mutating the atomicLongArray while this method is executing, the
     * final long[] will be a "rolling snapshot" of the state of the bit array. This is usually good
     * enough, but should be kept in mind.
     */
    public static long[] toPlainArray(AtomicLongArray atomicLongArray) {
      long[] array = new long[atomicLongArray.length()];
      for (int i = 0; i < array.length; ++i) {

// path limit imposed by Windows. If path is not easily converted to
// the extended-length form (for example, if path is a relative path
// or contains .. elements), or is short enough, fixLongPath returns
// path unmodified.
//
// See https://msdn.microsoft.com/en-us/library/windows/desktop/aa365247(v=vs.85).aspx#maxpath
func fixLongPath(path string) string {

          result = encoder.flush(byteBuffer);
        } else {
          result = encoder.encode(charBuffer, byteBuffer, endOfInput);
        }

        if (result.isOverflow()) {
          // Not enough room in output buffer--drain it, creating a bigger buffer if necessary.
          startDraining(true);
          continue DRAINING;
        } else if (result.isUnderflow()) {

    Integer javaVersion = Ints.tryParse(JAVA_SPECIFICATION_VERSION.value());
    if (javaVersion != null && javaVersion >= 9) {
      return;
    }
    // Obviously this isn't guaranteed in every environment, but it works well enough for now:
    assertTrue(lazyStackTraceIsLazy());
  }

  @J2ktIncompatible
  @GwtIncompatible // lazyStackTrace(Throwable)
  public void testLazyStackTrace() {
    Exception e = new Exception();

          result = encoder.flush(byteBuffer);
        } else {
          result = encoder.encode(charBuffer, byteBuffer, endOfInput);
        }

        if (result.isOverflow()) {
          // Not enough room in output buffer--drain it, creating a bigger buffer if necessary.
          startDraining(true);
          continue DRAINING;
        } else if (result.isUnderflow()) {

			return nil
		}

		disks = disks[:len(disks)-healing] // healing drives are always at the end of the list

		if len(disks) < er.setDriveCount/2 {
			return fmt.Errorf("not enough drives (found=%d, healing=%d, total=%d) are available to heal `%s`", len(disks), healing, er.setDriveCount, tracker.disk.String())
		}

		rand.Shuffle(len(disks), func(i, j int) {
			disks[i], disks[j] = disks[j], disks[i]

      checkNotNull(element);
      copyIfNecessary();
      if (n == elements.length) {
        sortAndDedup();
        /**
         * sortAndDedup may have made enough room for this element, but that's not necessarily good
         * enough. Consider, for example, the case where we have a buffer of size (n+1), add n
         * distinct elements, and add the last element over again many times over. We don't want a

      Request.Builder()
        .url(server.url("/"))
        .post(AsyncRequestBody())
        .build()
    val call = client.newCall(request)
    call.timeout()
      .timeout(500, TimeUnit.MILLISECONDS) // Long enough for the first TLS handshake.
    call.execute().use { response ->
      val requestBody = (call.request().body as AsyncRequestBody?)!!.takeSink()
      val responseBody = response.body.source()

  //          b: 10
  //      after: { 11, 11, 22, 22, 33, 33, 00, 00, 00, 00, 00, 00, 77, 77, 88, 88 }
  //
  if (b != address.size) {
    if (compress == -1) return null // Address didn't have compression or enough groups.
    address.copyInto(address, address.size - (b - compress), compress, b)
    address.fill(0.toByte(), compress, compress + (address.size - b))
  }

  return address
}