Question

I’m looking for a way to find how often I can divide a constant x by two (and not get a remainder) without using loops, recursion or the logarithm. Since this is the same problem as finding the index of the least significant non-zero bit, I am hoping that there is some way to use bitwise operations to do this. Unfortunately I wasn’t able to come up with it. Any ideas?

Background: I have a loop that doubles the counter on every iteration until it no longer divides the constant x. I want this loop to be unrolled, but the NVIDIA CUDA compiler isn’t smart enough to figure out the number of iterations, so I want to rewrite the loop in such a way that the number of iterations becomes more obvious to the compiler:

for(i=1; CONST & i == 0; i *= 2)
     bla(i);

should become something like

#define ITERATIONS missing_expr_with_CONST
for(i=0; i < ITERATIONS; i++)
     fasel(i);

Answer 1

This can be directly solved using this code for 32 bit numbers (I take no credit).

unsigned int v;  // find the number of trailing zeros in 32-bit v 
int r;           // result goes here
static const int MultiplyDeBruijnBitPosition[32] = 
{
  0, 1, 28, 2, 29, 14, 24, 3, 30, 22, 20, 15, 25, 17, 4, 8, 
  31, 27, 13, 23, 21, 19, 16, 7, 26, 12, 18, 6, 11, 5, 10, 9
};
r = MultiplyDeBruijnBitPosition[((uint32_t)((v & -v) * 0x077CB531U)) >> 27];