normalized[i] = parts[i].replaceFirst("^0+(?!$)", "");
            if (normalized[i].isEmpty()) {
                normalized[i] = "0";
            }
        }

        // Find longest sequence of consecutive zeros
        int longestStart = -1;
        int longestLength = 0;
        int currentStart = -1;
        int currentLength = 0;

        for (int i = 0; i < 8; i++) {

   * either it has exceeded the keep alive limit or the idle connections limit.
   *
   * Returns the duration in nanoseconds to sleep until the next scheduled call to this method.
   * Returns -1 if no further cleanups are required.
   */
  fun closeConnections(now: Long): Long {
    // Find the longest-idle connections in 2 categories:
    //

  return b == addressOffset + 4
}

/** Encodes an IPv6 address in canonical form according to RFC 5952. */
internal fun inet6AddressToAscii(address: ByteArray): String {
  // Go through the address looking for the longest run of 0s. Each group is 2-bytes.
  // A run must be longer than one group (section 4.2.2).
  // If there are multiple equal runs, the first one must be used (section 4.2.3).
  var longestRunOffset = -1

    for (n = len; n < size - n; n <<= 1) {
      System.arraycopy(array, 0, array, n, n);
    }
    System.arraycopy(array, 0, array, n, size - n);
    return new String(array);
  }

  /**
   * Returns the longest string {@code prefix} such that {@code a.toString().startsWith(prefix) &&
   * b.toString().startsWith(prefix)}, taking care not to split surrogate pairs. If {@code a} and
   * {@code b} have no common prefix, returns the empty string.

    for (n = len; n < size - n; n <<= 1) {
      System.arraycopy(array, 0, array, n, n);
    }
    System.arraycopy(array, 0, array, n, size - n);
    return new String(array);
  }

  /**
   * Returns the longest string {@code prefix} such that {@code a.toString().startsWith(prefix) &&
   * b.toString().startsWith(prefix)}, taking care not to split surrogate pairs. If {@code a} and
   * {@code b} have no common prefix, returns the empty string.

            // Leading zeros removed and consecutive zeros compressed
            assertEquals("2001:db8::1", method.invoke(null, "2001:0db8:0:0:0:0:0:1"));

            // Longest sequence of zeros compressed (3 zeros at end vs 1 zero at start)
            assertEquals("2001:db8:0:1::1", method.invoke(null, "2001:0db8:0:1:0:0:0:1"));

    // Start by looking for exact matches. We start at the leftmost label. For example, foo.bar.com
    // will look like: [foo, bar, com], [bar, com], [com]. The longest matching rule wins.
    var exactMatch: String? = null
    for (i in domainLabelsUtf8Bytes.indices) {
      val rule = publicSuffixList.bytes.binarySearch(domainLabelsUtf8Bytes, i)
      if (rule != null) {

                    }
                }
            }
        }

        if (dr == null && path != null) {
            /* We did not match a domain based root. Now try to match the
             * longest path in the list of stand-alone referrals.
             */
            if (referrals != null && now > referrals.expiration) {
                referrals = null;
            }
            if (referrals == null) {

/**
 * A bounded {@linkplain BlockingQueue blocking queue} backed by an array. This queue orders
 * elements FIFO (first-in-first-out). The head of the queue is that element that has been
 * on the queue the longest time. The tail of the queue is that element that has been on
 * the queue the shortest time. New elements are inserted at the tail of the queue, and the queue
 * retrieval operations obtain elements at the head of the queue.

            log.trace("No match for domain based root, checking standalone " + domain);
        }
        /*
         * We did not match a domain based root. Now try to match the
         * longest path in the list of stand-alone referrals.
         */

        CacheEntry<DfsReferralDataInternal> refs;
        synchronized (this.referralsLock) {
            refs = this.referrals;