Question no:3 bitmask output 3

#1

what is the precedence of the operators?

#2

hello @subhamjaiswal885

u can refer the list given here for precedance operator-> link

The answer should be log2( N & -N ) + 1 , if n is a positive number , so -n will be a negative number and negative number are represented in 2’s compliment ,which is ~n+1 , ~n has all the bits of n inverted due to which we get 0 in place of 1 and vice versa so when we add 1 to ~n then that 1 occupies the first rightmost 0 in (~n)(which is same as rightmost 1 value in n) and other bits have inverted values or zero values . so when we perform & operation we get only rightmost bit set in the function ,and then we perform logn (base2) to get corresponding index.

eg :- n = 1110010100
~n= 0001101011
~n+1=-n= 0001101100
(n&(-n))= 0000000100 = 4
log(4) =2
2+1=3 which is the answer

#3

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