Reverses the number of significant bits specified. - CSharp System

CSharp examples for System:Bit

  1. HOME
  2. CSharp
  3. System
  4. Bit

Description

Reverses the number of significant bits specified.

Demo Code


using System;/*from w  ww.  jav  a 2  s  . c om*/

public class Main{
        /// <summary>
      /// Reverses the number of significant bits specified.
      /// </summary>
      /// <param name="value">The value to reverse.</param>
      /// <param name="significantBitCount">The number of significant bits to reverse.</param>
      /// <returns>The <paramref name="value"/> with its <paramref name="significantBitCount"/> reversed.</returns>
      /// <example>
      /// Example, if the input is 376, with significantBitCount=11, the output is 244 (decimal, base 10).
      /// 
      /// 376 = 00000101111000
      /// 244 = 00000011110100
      /// 
      /// Example, if the input is 900, with significantBitCount=11, the output is 270.
      /// 
      /// 900 = 00001110000100
      /// 270 = 00000100001110
      /// 
      /// Example, if the input is 900, with significantBitCount=12, the output is 540.
      /// 
      /// 900 = 00001110000100
      /// 540 = 00001000011100
      /// 
      /// Example, if the input is 154, with significantBitCount=4, the output is 5.
      /// 
      /// 154 = 00000010011010
      /// 005 = 00000000000101
      /// </example>
      public static long Reverse(long value, int significantBitCount)
      {
         ulong n = (ulong) value << (64 - significantBitCount);
         n = n >> 32 | n << 32;
         n = n >> 0xf & 0x0000ffff | n << 0xf & 0xffff0000;
         n = n >> 0x8 & 0x00ff00ff | n << 0x8 & 0xff00ff00;
         n = n >> 0x4 & 0x0f0f0f0f | n << 0x4 & 0xf0f0f0f0;
         n = n >> 0x2 & 0x33333333 | n << 0x2 & 0xcccccccc;
         n = n >> 0x1 & 0x55555555 | n << 0x1 & 0xaaaaaaaa;

         return (long) n | value & (-1 << significantBitCount);
      }
}

Related Tutorials