It is very useful. Some common questions like:
Many tricks are based on 
1. sign = v >> (sizeof(int) * 8 - 1); // if v < v =" ~v" v="1100"> 1000

(1) abs() 
     cwd  //get the sign of [ax] into dx: -1 when <0 ax =" [ax]">0 else -([ax])-1    
     sub ax,dx   // ax = [ax] when >0 else -([ax])-1-(-1);

(2) power of 2
f = !(v & (v - 1)) && v;

(3) Counting bit set
Brian Kernighan's way
  unsigned int v; // count the number of bits set in v
  unsigned int c; // c accumulates the total bits set in v
  for (c = 0; v; c++)
  {
    v &= v - 1; // clear the least significant bit set
  }
parallel sideway addition
  unsigned int v; // count bits set in this (32-bit value)
  unsigned int c; // store the total here
  const int S[] = {1, 2, 4, 8, 16}; // Magic Binary Numbers
  const int B[] = {0x55555555, 0x33333333, 0x0F0F0F0F, 0x00FF00FF, 0x0000FFFF};

  c = v;
  c = ((c >> S[0]) & B[0]) + (c&B[0]);
  //add adjacent one bits 1,1->10 1,0->01 0,1->01 0,0->00
  c = ((c >> S[1]) & B[1]) + (c&B[1]); //
  //add adjacent two bits 00,00->0000 00,01->0001 01,00->0001 etc...
  c = ((c >> S[2]) & B[2]) + (c&B[2]);
  //add adjacent four bits 0000,0000->00000000, etc
  c = ((c >> S[3]) & B[3]) + (c&B[3]);
  //add adjacent eight bits
  c = ((c >> S[4]) & B[4]) + (c&B[4]);
  //add adjacent sixteen bits


(4) increament without "+" or "-"
(~a) * (~0)
[equiv] (-a-1) * (-1) = a+1
-0 = ~0 + 1 ==> ~0 = -1

(5) Modulo 3 without equation X-X/3
Given x = v>>2 we have
v= 4*x + v&3 = 3*x + x + v&3
with v mod 3 == (3*x + x + v&3) mod 3 == (x+v&3) mod 3 == (v>>2+v&3) mod 3
so we have
int Modulo3(const int x)
{
  int v = x;
  while (v > 3)v = v>>2 + v&3;
  if (v>2) v=v-3;
  return v;
}
the same idea fits Modulo 5
int Modulo5(const int x)
{
  int v = x;
  while (v>15)
  {
    v = v >>4 + v&15;
    // given x = v>>4;
    // v mod 5 == (16*x+v&15) mod 5
    // == (15*x + x + v&15) mod 5
    // == (x+v&15) mod 5 == ( v>>4+v&15) mod 5
  }
  while (v>4)
  {
    v = v - 5;
  }
  return v;
}

- posted by Pop @ 10:20 PM