this.bitCount = LongAddables.create();
    }

    // Used by serialization
    LockFreeBitArray(long[] data) {
      checkArgument(data.length > 0, "data length is zero!");
      this.data = new AtomicLongArray(data);
      this.bitCount = LongAddables.create();
      long bitCount = 0;
      for (long value : data) {
        bitCount += Long.bitCount(value);
      }

   *
   * @since 22.0
   */
  public long approximateElementCount() {
    long bitSize = bits.bitSize();
    long bitCount = bits.bitCount();

    /*
     * Each insertion is expected to reduce the # of clear bits by a factor of
     * `numHashFunctions/bitSize`. So, after n insertions, expected bitCount is `bitSize * (1 - (1 -
     * numHashFunctions/bitSize)^n)`. Solving that for n, and approximating `ln x` as `x - 1` when x

            assertEquals(1, Integer.bitCount(DcerpcConstants.DCERPC_CONC_MPX), "DCERPC_CONC_MPX should be a single bit");
            assertEquals(1, Integer.bitCount(DcerpcConstants.DCERPC_DID_NOT_EXECUTE), "DCERPC_DID_NOT_EXECUTE should be a single bit");
            assertEquals(1, Integer.bitCount(DcerpcConstants.DCERPC_MAYBE), "DCERPC_MAYBE should be a single bit");

        key8[6] = (byte) (key[5] << 2 | (key[6] & 0xFF) >>> 6);
        key8[7] = (byte) (key[6] << 1);
        for (int i = 0; i < key8.length; i++) {
            key8[i] ^= Integer.bitCount(key8[i] ^ 1) & 1;
        }
        return key8;
    }

    /**
     * Default provider is BouncyCastleProvider.
     * For registering custom provider
     * @see jcifs.util.Crypto#initProvider(Provider)

    checkPositive("x", x);
    return Integer.highestOneBit(x);
  }

  /**
   * Returns {@code true} if {@code x} represents a power of two.
   *
   * <p>This differs from {@code Integer.bitCount(x) == 1}, because {@code
   * Integer.bitCount(Integer.MIN_VALUE) == 1}, but {@link Integer#MIN_VALUE} is not a power of two.
   */
  // Whenever both tests are cheap and functional, it's faster to use &, | instead of &&, ||

  public void testIsPowerOfTwo() {
    for (int x : ALL_INTEGER_CANDIDATES) {
      // Checks for a single bit set.
      BigInteger bigX = BigInteger.valueOf(x);
      boolean expected = (bigX.signum() > 0) && (bigX.bitCount() == 1);
      assertEquals(expected, IntMath.isPowerOfTwo(x));
    }
  }

  public void testLog2ZeroAlwaysThrows() {
    for (RoundingMode mode : ALL_ROUNDING_MODES) {

  public void testIsPowerOfTwo() {
    for (int x : ALL_INTEGER_CANDIDATES) {
      // Checks for a single bit set.
      BigInteger bigX = BigInteger.valueOf(x);
      boolean expected = (bigX.signum() > 0) && (bigX.bitCount() == 1);
      assertEquals(expected, IntMath.isPowerOfTwo(x));
    }
  }

  public void testLog2ZeroAlwaysThrows() {
    for (RoundingMode mode : ALL_ROUNDING_MODES) {

    return 1L << ((Long.SIZE - 1) - Long.numberOfLeadingZeros(x));
  }

  /**
   * Returns {@code true} if {@code x} represents a power of two.
   *
   * <p>This differs from {@code Long.bitCount(x) == 1}, because {@code
   * Long.bitCount(Long.MIN_VALUE) == 1}, but {@link Long#MIN_VALUE} is not a power of two.
   */
  // Whenever both tests are cheap and functional, it's faster to use &, | instead of &&, ||

  public void testIsPowerOfTwo() {
    for (long x : ALL_LONG_CANDIDATES) {
      // Checks for a single bit set.
      BigInteger bigX = BigInteger.valueOf(x);
      boolean expected = (bigX.signum() > 0) && (bigX.bitCount() == 1);
      assertEquals(expected, LongMath.isPowerOfTwo(x));
    }
  }

  public void testLog2ZeroAlwaysThrows() {
    for (RoundingMode mode : ALL_ROUNDING_MODES) {

          }
          remainingSetBits &= ~(1 << index);
          return elements.get(index);
        }
      };
    }

    @Override
    public int size() {
      return Integer.bitCount(mask);
    }

    @Override
    public boolean contains(@Nullable Object o) {
      Integer index = inputSet.get(o);
      return index != null && (mask & (1 << index)) != 0;
    }
  }