}
      }

      long buffer = 0;
      int shift = 0;
      for (; i < utf16Length; i++) {
        char c = input.charAt(i);
        if (c < 0x80) {
          buffer |= (long) c << shift;
          shift += 8;
          len++;
        } else if (c < 0x800) {
          buffer |= charToTwoUtf8Bytes(c) << shift;
          shift += 16;
          len += 2;

      if ((radix & (radix - 1)) == 0) {
        // Radix is a power of two so we can avoid division.
        int shift = Integer.numberOfTrailingZeros(radix);
        int mask = radix - 1;
        do {
          buf[--i] = Character.forDigit(((int) x) & mask, radix);
          x >>>= shift;
        } while (x != 0);
      } else {
        // Separate off the last digit using unsigned division. That will leave

   */
  public static final byte CR = 13;

  /**
   * Shift Out: A control character indicating that the code combinations which follow shall be
   * interpreted as outside of the character set of the standard code table until a Shift In
   * character is reached.
   *
   * @since 8.0
   */
  public static final byte SO = 14;

  /**

        int remainingPowersOf2 = 32;
        do {
          int shift = min(remainingPowersOf2, Long.numberOfLeadingZeros(a));
          // shift is either the number of powers of 2 left to multiply a by, or the biggest shift
          // possible while keeping a in an unsigned long.
          a = UnsignedLongs.remainder(a << shift, m);
          remainingPowersOf2 -= shift;
        } while (remainingPowersOf2 > 0);
        return a;

     * top SIGNIFICAND_BITS + 1.
     *
     * It helps to consider the real number signif = absX * 2^(SIGNIFICAND_BITS - exponent).
     */
    int shift = exponent - SIGNIFICAND_BITS - 1;
    long twiceSignifFloor = absX.shiftRight(shift).longValue();
    long signifFloor = twiceSignifFloor >> 1;
    signifFloor &= SIGNIFICAND_MASK; // remove the implied bit

    /*

 * to unsigned arithmetic in all cases except:
 *
 * <ul>
 *   <li>comparisons (signed values can be negative)
 *   <li>division (avoided here)
 *   <li>shifting (right shift must be unsigned)
 * </ul>
 *
 * @author ******@****.*** (Kyle Maddison)
 * @author ******@****.*** (Geoff Pike)
 */
@ElementTypesAreNonnullByDefault

    long result = 0;
    // Due to the way we shift, we can stop iterating once we've run out of data, the rest
    // of the result already being filled with zeros.

    // This loop is critical to performance, so please check HashBenchmark if altering it.
    int limit = Math.min(length, 8);
    for (int i = 0; i < limit; i++) {
      // Shift value left while iterating logically through the array.

    // 1) array has space available on right-hand side (between limit and capacity)
    // 2) array has space available on left-hand side (before position)
    // 3) array has no space available
    //
    // In case 2 we shift the existing chars to the left, and in case 3 we create a bigger
    // array, then they both become case 1.

    if (availableCapacity(charBuffer) == 0) {
      if (charBuffer.position() > 0) {

             * little-endian. Long.numberOfTrailingZeros(diff) tells us the least significant
             * nonzero bit, and zeroing out the first three bits of L.nTZ gives us the shift to get
             * that least significant nonzero byte.
             */
            int n = Long.numberOfTrailingZeros(lw ^ rw) & ~0x7;

      return (double) value;
    }
    // The top bit is set, which means that the double value is going to come from the top 53 bits.
    // So we can ignore the bottom 11, except for rounding. We can unsigned-shift right 1, aka
    // unsigned-divide by 2, and convert that. Then we'll get exactly half of the desired double
    // value. But in the specific case where the bottom two bits of the original number are 01, we