Divide a signed integer by a power of 2

Question

I\'m working on a way to divide a signed integer by a power of 2 using only binary operators (<< >> + ^ ~ & | !), and the result has to be round toward 0. I came a

Answer 1

OP's reference is of a C# code and so many subtle differences that cause it to be bad code with C, as this post is tagged.

int is not necessarily 32-bits so using a magic number of 32 does not make for a robust solution.

In particular (1 << n) + ~0 results in implementation defined behavior when n causes a bit to be shifted into the sign place. Not good coding.

Restricting code to only using "binary" operators << >> + ^ ~ & | ! encourages a coder to assume things about int which is not portable nor compliant with the C spec. So OP's posted code does not "work" in general, although may work in many common implementations.

OP code fails when int is not 2's complement, not uses the range [-2147483648 .. 2147483647] or when 1 << n uses implementation behavior that is not as expected.

// weak code
int divideByPowerOf2(int x, int n) {
  return (x + ((x >> 31) & ((1 << n) + ~0))) >> n;
}

A simple alternative, assuming long long exceeds the range of int follows. I doubt this meets some corner of OP's goals, but OP's given goals encourages non-robust coding.

int divideByPowerOf2(int x, int n) {
  long long ill = x;
  if (x < 0) ill = -ill;
  while (n--) ill >>= 1;
  if (x < 0) ill = -ill;
  return (int) ill;
}