Test if a bitboard have only one bit set to 1 [duplicate]

Question

This question already has answers here:

How to check if a number is a power of 2 (25 answers)

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I have a bitboard and I want to check in C if there is ONLY one bit set to 1.

#include <stdint.h>
typedef uint64_t bboard;
bboard b = 0x0000000000000010;
if (only_one_bit_set_to_one (b)) // in this example expected true
  // do something...

Any idea to write the function int only_one_bit_set_to_one (bboard b)?

Answer 1

Sure, it's easy:

int only_one_bit_set_to_one (bboard b)
{
    return b && !(b & (b-1));
}

Say b has any bits set, the least significant is bit number k. Then b-1 has the same bits as b for indices above k, a 0-bit in place k and 1-bits in the less significant places, so the bitwise and removes the least significant set bit from b. If b had only one bit set, the result becomes 0, if b had more bits set, the result is nonzero.

Answer 2

This may be a bit naive, but I'd loop from 0 to 63, clear the relevant bit, and see if the result is 0:

if (b != 0) {
    for (i = 0; i < 64; ++i) {
        if (b & ~(1 << i)) == 0) {
            return 1;
        }
    }
    return 0;
}

it's nowhere near as clever as the other posted answers, but it has the advantage of being easy to understand.