Search code examples
optimizationbit-manipulationbitflags

Determine Position of Most Signifiacntly Set Bit in a Byte

I have a byte I am using to store bit flags. I need to compute the position of the most significant set bit in the byte.

Example Byte: 00101101 => 6 is the position of the most significant set bit

Compact Hex Mapping:

[0x00]      => 0x00
[0x01]      => 0x01
[0x02,0x03] => 0x02
[0x04,0x07] => 0x03
[0x08,0x0F] => 0x04
[0x10,0x1F] => 0x05
[0x20,0x3F] => 0x06
[0x40,0x7F] => 0x07
[0x80,0xFF] => 0x08

TestCase in C:

#include <stdio.h>

unsigned char check(unsigned char b) {
  unsigned char c = 0x08;
  unsigned char m = 0x80;
  do {
    if(m&b) { return  c; }
    else    { c -= 0x01; }
  } while(m>>=1);
  return 0; //never reached
}
int main() {
  unsigned char input[256] = {
    0x00,0x01,0x02,0x03,0x04,0x05,0x06,0x07,0x08,0x09,0x0a,0x0b,0x0c,0x0d,0x0e,0x0f,
    0x10,0x11,0x12,0x13,0x14,0x15,0x16,0x17,0x18,0x19,0x1a,0x1b,0x1c,0x1d,0x1e,0x1f,
    0x20,0x21,0x22,0x23,0x24,0x25,0x26,0x27,0x28,0x29,0x2a,0x2b,0x2c,0x2d,0x2e,0x2f,
    0x30,0x31,0x32,0x33,0x34,0x35,0x36,0x37,0x38,0x39,0x3a,0x3b,0x3c,0x3d,0x3e,0x3f,
    0x40,0x41,0x42,0x43,0x44,0x45,0x46,0x47,0x48,0x49,0x4a,0x4b,0x4c,0x4d,0x4e,0x4f,
    0x50,0x51,0x52,0x53,0x54,0x55,0x56,0x57,0x58,0x59,0x5a,0x5b,0x5c,0x5d,0x5e,0x5f,
    0x60,0x61,0x62,0x63,0x64,0x65,0x66,0x67,0x68,0x69,0x6a,0x6b,0x6c,0x6d,0x6e,0x6f,
    0x70,0x71,0x72,0x73,0x74,0x75,0x76,0x77,0x78,0x79,0x7a,0x7b,0x7c,0x7d,0x7e,0x7f,
    0x80,0x81,0x82,0x83,0x84,0x85,0x86,0x87,0x88,0x89,0x8a,0x8b,0x8c,0x8d,0x8e,0x8f,
    0x90,0x91,0x92,0x93,0x94,0x95,0x96,0x97,0x98,0x99,0x9a,0x9b,0x9c,0x9d,0x9e,0x9f,
    0xa0,0xa1,0xa2,0xa3,0xa4,0xa5,0xa6,0xa7,0xa8,0xa9,0xaa,0xab,0xac,0xad,0xae,0xaf,
    0xb0,0xb1,0xb2,0xb3,0xb4,0xb5,0xb6,0xb7,0xb8,0xb9,0xba,0xbb,0xbc,0xbd,0xbe,0xbf,
    0xc0,0xc1,0xc2,0xc3,0xc4,0xc5,0xc6,0xc7,0xc8,0xc9,0xca,0xcb,0xcc,0xcd,0xce,0xcf,
    0xd0,0xd1,0xd2,0xd3,0xd4,0xd5,0xd6,0xd7,0xd8,0xd9,0xda,0xdb,0xdc,0xdd,0xde,0xdf,
    0xe0,0xe1,0xe2,0xe3,0xe4,0xe5,0xe6,0xe7,0xe8,0xe9,0xea,0xeb,0xec,0xed,0xee,0xef,
    0xf0,0xf1,0xf2,0xf3,0xf4,0xf5,0xf6,0xf7,0xf8,0xf9,0xfa,0xfb,0xfc,0xfd,0xfe,0xff };

  unsigned char truth[256] = {
    0x00,0x01,0x02,0x02,0x03,0x03,0x03,0x03,0x04,0x04,0x04,0x04,0x04,0x04,0x04,0x04, 
    0x05,0x05,0x05,0x05,0x05,0x05,0x05,0x05,0x05,0x05,0x05,0x05,0x05,0x05,0x05,0x05, 
    0x06,0x06,0x06,0x06,0x06,0x06,0x06,0x06,0x06,0x06,0x06,0x06,0x06,0x06,0x06,0x06, 
    0x06,0x06,0x06,0x06,0x06,0x06,0x06,0x06,0x06,0x06,0x06,0x06,0x06,0x06,0x06,0x06, 
    0x07,0x07,0x07,0x07,0x07,0x07,0x07,0x07,0x07,0x07,0x07,0x07,0x07,0x07,0x07,0x07, 
    0x07,0x07,0x07,0x07,0x07,0x07,0x07,0x07,0x07,0x07,0x07,0x07,0x07,0x07,0x07,0x07, 
    0x07,0x07,0x07,0x07,0x07,0x07,0x07,0x07,0x07,0x07,0x07,0x07,0x07,0x07,0x07,0x07, 
    0x07,0x07,0x07,0x07,0x07,0x07,0x07,0x07,0x07,0x07,0x07,0x07,0x07,0x07,0x07,0x07, 
    0x08,0x08,0x08,0x08,0x08,0x08,0x08,0x08,0x08,0x08,0x08,0x08,0x08,0x08,0x08,0x08,
    0x08,0x08,0x08,0x08,0x08,0x08,0x08,0x08,0x08,0x08,0x08,0x08,0x08,0x08,0x08,0x08,
    0x08,0x08,0x08,0x08,0x08,0x08,0x08,0x08,0x08,0x08,0x08,0x08,0x08,0x08,0x08,0x08,
    0x08,0x08,0x08,0x08,0x08,0x08,0x08,0x08,0x08,0x08,0x08,0x08,0x08,0x08,0x08,0x08,
    0x08,0x08,0x08,0x08,0x08,0x08,0x08,0x08,0x08,0x08,0x08,0x08,0x08,0x08,0x08,0x08,
    0x08,0x08,0x08,0x08,0x08,0x08,0x08,0x08,0x08,0x08,0x08,0x08,0x08,0x08,0x08,0x08,
    0x08,0x08,0x08,0x08,0x08,0x08,0x08,0x08,0x08,0x08,0x08,0x08,0x08,0x08,0x08,0x08,
    0x08,0x08,0x08,0x08,0x08,0x08,0x08,0x08,0x08,0x08,0x08,0x08,0x08,0x08,0x08,0x08};

  int i,r;
  int f = 0;
  for(i=0; i<256; ++i) {
    r=check(input[i]);
    if(r !=(truth[i])) {
      printf("failed %d : 0x%x : %d\n",i,0x000000FF & ((int)input[i]),r);
      f += 1;
    }
  }
  if(!f) { printf("passed all\n");  }
  else   { printf("failed %d\n",f); }
  return 0;
}

I would like to simplify my check() function to not involve looping (or branching preferably). Is there a bit twiddling hack or hashed lookup table solution to compute the position of the most significant set bit in a byte?

Solution

  • Your question is about an efficient way to compute log2 of a value. And because you seem to want a solution that is not limited to the C language I have been slightly lazy and tweaked some C# code I have.

    You want to compute log2(x) + 1 and for x = 0 (where log2 is undefined) you define the result as 0 (e.g. you create a special case where log2(0) = -1).

    static readonly Byte[] multiplyDeBruijnBitPosition = new Byte[] {
  7, 2, 3, 4,
  6, 1, 5, 0
};

public static Byte Log2Plus1(Byte value) {
  if (value == 0)
    return 0;

  var roundedValue = value;
  roundedValue |= (Byte) (roundedValue >> 1);
  roundedValue |= (Byte) (roundedValue >> 2);
  roundedValue |= (Byte) (roundedValue >> 4);
  var log2 = multiplyDeBruijnBitPosition[((Byte) (roundedValue*0xE3)) >> 5];
  return (Byte) (log2 + 1);
}

    This bit twiddling hack is taken from Find the log base 2 of an N-bit integer in O(lg(N)) operations with multiply and lookup where you can see the equivalent C source code for 32 bit values. This code has been adapted to work on 8 bit values.

    However, you may be able to use an operation that gives you the result using a very efficient built-in function (on many CPU's a single instruction like the Bit Scan Reverse is used). An answer to the question Bit twiddling: which bit is set? has some information about this. A quote from the answer provides one possible reason why there is low level support for solving this problem:

    Things like this are the core of many O(1) algorithms such as kernel schedulers which need to find the first non-empty queue signified by an array of bits.