I have an integer of type long long that I want to convert to a new integer after removing the trailing zeros of that integer that are present in its binary representation.
Here is a brute force approach:
long long remove_trailing_zeroes(long long v) {
if (v != 0) {
while ((v & 1) == 0)
v /= 2;
}
return v;
}
Here is a direct approach for unsigned numbers, but the division might be more costly than the above iteration:
unsigned long long remove_trailing_zeroes(unsigned long long v) {
if (v != 0) {
// v and (v - 1) differ only in the trailing 0 bits plus 1
// shifting v ^ (v - 1) right by 1 and adding 1 gives the power of 2
// by which to divide v to remove all trailing 0 bits
v /= (((v ^ (v - 1)) >> 1) + 1);
}
return v;
}
harold suggested this simplification:
unsigned long long remove_trailing_zeroes(unsigned long long v) {
if (v != 0) {
// `-v`, which is `(~v + 1)` has all bits flipped except the least
// significant 1 bit.
// dividing v by `-v & v` shifts all trailing zero bits out,
v /= -v & v;
}
return v;
}
Which can be simplified as a single expression:
unsigned long long remove_trailing_zeroes(unsigned long long v) {
return v ? v / (-v & v) : v;
}
To avoid the division, you could count the number of bits in v ^ (v - 1)
with an efficient method and shift v
right by one less than this number. This would work for 0
as well so you would get branchless code.
You can find other methods in the fascinating word of Bit Twiddling Hacks.