Question

I’m trying to do a bit reversal, I understand the straight forward implementation, but for performance purposes I need to reverse bits by building a lookup table and using that table to find the bit reversals, So my program will build a look up table for bitsize N, and then it will use the table to find the bit reversal for given bits.

for example, if bitSize = 9 and num = 6 (00000110)

reverse(num,bitSize) will return 192 which is 11000000

int lookup(int num, int bitSize)
{
    return table[bitSize][num]; // == reverse(num,bitSize);
}

I think this is how the lookup function should look like, I’ve been told that it’s possible to build the table but I don’t have any idea how to, can someone explain how to build this table?

I just want to clarify that I’m looking for a way to build this table for given bitSize, not just for 32-bit, otherwise I would have used this method:
http://graphics.stanford.edu/~seander/bithacks.html#BitReverseTable

Thank you for your help,

Edit:
Both solutions work but kmkaplan’s solution is more efficient, thanks to j_random_hacker’s memory optimisation.

Answer 1

#include <stdio.h>
#include <stdlib.h>
int
main(int argc, char **argv)
{
  const int bitSize = atoi(argv[1]);

  printf("static unsigned int table[] = {");
  unsigned int i;
  for (i = 0; i < 1 << bitSize; i++) {
    if ((i & 7) == 0)
      printf("\n");
    unsigned int v = i;
    unsigned int r = 0;
    int s = bitSize;
    while (s--) {
      r <<= 1;
      r |= v & 1;
      v >>= 1;
    }
    printf(" 0x%x,", r);
  }
  printf("\n};\n"
         "unsigned int lookup(int num, int bitSize)\n"
         "{\n"
         "        return table[num] >> (%d - bitSize);\n"
         "}\n",
         bitSize
         );

  return 0;
}

edit: implement j_random_hacker memory optimisation.