Dividing by arbitrary numbers using shifting operators

Question

How can you divide a number n for example by 24 using shifting operators and additions?

( n % 24 == 0 )

Answer 1

The pencil-and-paper algorithm for the division uses only "shifts" (base 10 shifts) and subtractions. You can do exactly the same thing in base 2.

Sorry I can't find a link for the algorithm, but you should have learnt it as a child.

EDIT: actually since additions are cheap, you don't have to try to extract the correct digits one by one, so you can simplify the algorithm a bit...

Assuming positive dividend and divisor...

Take the power of two immediately larger than the divisor (here, 32).

It is easy to divide your number by this power of two. Say the division produces k1 . Subtract k1*24 from the number (call the rest r1 ) and iterate...

When you have obtained k1 , k2 , ... kn numbers and the rest rn no longer contains 32, do a last check to see if rn contains 24.

The result of the division is k1+k2+...+kn (+1 if 24 fits in rn) .

Answer 2

This works by first finding the highest bit of the result, and then working back.

int div24(int value) {
  // Find the smallest value of n such that (24<<n) > value
  int tmp = 24;
  for (int n = 0; tmp < value; ++n)
    tmp <<= 1;
  // Now start working backwards to find bits of the result. This is O(i).
  int result = 0;
  while(value != 0) {
    tmp >>= 1;
    result <<= 1;
    if (value > tmp) { // Found one bit.
       value -= tmp; // Subtract (24<<i)
       result++;
    }
  }
  return result;
}

Example:

Value = 120 :  n = 2
Step 0: tmp = 96, result = 0, value = 24, result = 1
Step 1: tmp = 48, result = 2
Step 2: tmp = 24, result = 4, value = 0, result = 5

Answer 3

int
div24(int value) {
   int result = 0;
   while (value >= 24) {       
      int accum = 24;
      int tmp = 1;    
      while (accum + accum <= value) {
         accum += accum;
         tmp += tmp;
      }
      value -= accum;     
      result += tmp;
   }   
   return result;
}

Answer 4

尽管毫无疑问，对于24来说，这是一个聪明的技巧，但使用移位运算符（和/或减法）仅对2的幂才有意义。即使那样，它也更可能使读者感到困惑，而不是用现代的方法生成更有效的代码。编译器。