Question

Are there any efficient bitwise operations I can do to get the number of set bits that an integer ends with? For example 11₁₀ = 1011₂ would be two trailing 1 bits. 8₁₀ = 1000₂ would be 0 trailing 1 bits.

Is there a better algorithm for this than a linear search? I’m implementing a randomized skip list and using random numbers to determine the maximum level of an element when inserting it. I am dealing with 32 bit integers in C++.

Edit: assembler is out of the question, I’m interested in a pure C++ solution.

Answer 1

Taking the answer from Ignacio Vazquez-Abrams and completing it with the count rather than a table:

b = ~i & (i+1);   // this gives a 1 to the left of the trailing 1's
b--;              // this gets us just the trailing 1's that need counting
b = (b & 0x55555555) + ((b>>1) & 0x55555555);  // 2 bit sums of 1 bit numbers
b = (b & 0x33333333) + ((b>>2) & 0x33333333);  // 4 bit sums of 2 bit numbers
b = (b & 0x0f0f0f0f) + ((b>>4) & 0x0f0f0f0f);  // 8 bit sums of 4 bit numbers
b = (b & 0x00ff00ff) + ((b>>8) & 0x00ff00ff);  // 16 bit sums of 8 bit numbers
b = (b & 0x0000ffff) + ((b>>16) & 0x0000ffff); // sum of 16 bit numbers

at the end b will contain the count of 1’s (the masks, adding and shifting count the 1’s).
Unless I goofed of course. Test before use.