if (tableSize <= BYTE_MAX_SIZE) {
      /*
       * Use 8 bits per entry. The value is unsigned to allow use up to a size of 2^8.
       *
       * The absent indicator of -1 signed becomes 2^8 - 1 unsigned, which reduces the actual max
       * size to 2^8 - 1. However, due to a load factor < 1 the limit is never approached.
       */
      byte[] hashTable = new byte[tableSize];

        E e = q.poll(deadline - System.nanoTime(), TimeUnit.NANOSECONDS);
        if (e == null) {
          break; // we already waited enough, and there are no more elements in sight
        }
        buffer.add(e);
        added++;
      }
    }
    return added;
  }

  /**
   * Drains the queue as {@linkplain #drain(BlockingQueue, Collection, int, long, TimeUnit)}, but

public class LongMathTest extends TestCase {
  @SuppressWarnings("ConstantOverflow")
  public void testMaxSignedPowerOfTwo() {
    assertTrue(LongMath.isPowerOfTwo(LongMath.MAX_SIGNED_POWER_OF_TWO));
    assertFalse(LongMath.isPowerOfTwo(LongMath.MAX_SIGNED_POWER_OF_TWO * 2));
  }

  public void testCeilingPowerOfTwo() {
    for (long x : POSITIVE_LONG_CANDIDATES) {

   * serialization.
   *
   * <p>Use {@linkplain #readFrom(InputStream, Funnel)} to reconstruct the written BloomFilter.
   */
  public void writeTo(OutputStream out) throws IOException {
    // Serial form:
    // 1 signed byte for the strategy
    // 1 unsigned byte for the number of hash functions
    // 1 big endian int, the number of longs in our bitset
    // N big endian longs of our bitset

public class LongMathTest extends TestCase {
  @SuppressWarnings("ConstantOverflow")
  public void testMaxSignedPowerOfTwo() {
    assertTrue(LongMath.isPowerOfTwo(LongMath.MAX_SIGNED_POWER_OF_TWO));
    assertFalse(LongMath.isPowerOfTwo(LongMath.MAX_SIGNED_POWER_OF_TWO * 2));
  }

  public void testCeilingPowerOfTwo() {
    for (long x : POSITIVE_LONG_CANDIDATES) {