Unset the rightmost set bit [duplicate]

Answer 1

unsigned int clr_rm_set_bit(unsigned int n)
{
    unsigned int mask = 1;
    while(n & mask) {
        mask <<= 1;
    }
    return n & ~mask;
}

Answer 2

Try n & (n-1) where & is bitwise AND

n = 7
n - 1 =6

n & (n-1)=> 0 1 1 1   (7)
          & 0 1 1 0   (6)
           --------- 
            0 1 1 0  (done!)

EDIT (in response to the comment given by Forest)

n = 6 
n - 1 = 5

n & (n-1)=> 0 1 1 0   (6)
          & 0 1 0 1   (5)
           --------- 
            0 1 0 0  (done!)

Answer 3

Your question is unclear.

If you just want to unset bit 0, here are some methods (with slight variations in behavior depending on your types involved):

x &= -2;
x &= ~1;
x -= (x&1);

If you want to unset the lowest bit among the bits that are set, here are some ways:

x &= x-1;
x -= (x&-x);

Note that x&-x is equal to the lowest bit of x, at least when x is unsigned or twos complement. If you want to do any bit arithmetic like this, you should use only unsigned types, since signed types have implementation-defined behavior under bitwise operations.