Bitwise operator for simply flipping all bits in an integer?

Question

I have to flip all bits in a binary representation of an integer. Given:

10101 

The output should be

01010 

What is the bitwise operator to accomplish this when used with an integer? For example, if I were writing a method like int flipBits(int n);, what would go in the body? I need to flip only what's already present in the number, not all 32 bits in the integer.

Answer 1

The ~ unary operator is bitwise negation. If you need fewer bits than what fits in an int then you'll need to mask it with & after the fact.

Answer 2

Simply use the bitwise not operator ~.

int flipBits(int n) {     return ~n; } 

To use the k least significant bits, convert it to the right mask.
(I assume you want at least 1 bit of course, that's why mask starts at 1)

int flipBits(int n, int k) {     int mask = 1;     for (int i = 1; i < k; ++i)         mask |= mask << 1;      return ~n & mask; } 

As suggested by Lưu Vĩnh Phúc, one can create the mask as (1 << k) - 1 instead of using a loop.

int flipBits2(int n, int k) {     int mask = (1 << k) - 1;     return ~n & mask; }