How to get position of right most set bit in C

This answer Unset the rightmost set bit tells both how to get and unset rightmost set bit for an unsigned integer or signed integer represented as two's complement.

get rightmost set bit,

x & -x
// or
x & (~x + 1)

unset rightmost set bit,

x &= x - 1
// or
x -= x & -x  // rhs is rightmost set bit

why it works

x:                     leading bits  1  all 0
~x:           reversed leading bits  0  all 1
~x + 1 or -x: reversed leading bits  1  all 0
x & -x:                       all 0  1  all 0

eg, let x = 112, and choose 8-bit for simplicity, though the idea is same for all size of integer.

// example for get rightmost set bit
x:             01110000
~x:            10001111
-x or ~x + 1:  10010000
x & -x:        00010000

// example for unset rightmost set bit
x:             01110000
x-1:           01101111
x & (x-1):     01100000

Finding the (0-based) index of the least significant set bit is equivalent to counting how many trailing zeros a given integer has. Depending on your compiler there are builtin functions for this, for example gcc and clang support __builtin_ctz. For MSVC you would need to implement your own version, this answer to a different question shows a solution making use of MSVC intrinsics.

Given that you are looking for the 1-based index, you simply need to add 1 to ctz's result in order to achieve what you want.

int a = 12;
int least_bit = __builtin_ctz(a) + 1; // least_bit = 3

Note that this operation is undefined if a == 0. Furthermore there exist __builtin_ctzl and __builtin_ctzll which you should use if you are working with long and long long instead of int.