Invert bits of binary representation of number

Question

This is the code I came up with.

I added comments to make the solution more verbose.

int findComplement(int num) {
 // b is the answer which will be returned
 int b = 0;
 // One bit will be taken at a time from num, will be inverted and stored in n for adding to result
 int n = 0;
 // k will be used to shift bit to be inserted in correct position
 int k = 0;
 while(num){
 // Invert bit of current number
 n = !(num & 1);
 // Shift the given number one bit right to accesss next bit in next iteration
 num = num >>1 ;
 // Add the inverted bit after shifting
 b = b + (n<<k);
 // Increment the number by which to shift next bit
 k++;
 }
 return b;
}

Is there any redundant statment in my code which can be removed? Or any other better logic to invert bits of a given integer

• Martin R: Are you re-inventing the binary not operator (~)?
• KshitijV97: I don't want to sound dumb, But honestly, I did not know that ~ operator existed which inverts all bits of a given integer.
• AJNeufeld: Many easy ways. ~num or -1 - num, or 0xFFFFFFFF - num, or 0xFFFFFFFF ^ num or (-1) ^ num. Doing it one bit at a time is most definitely the hard way.

Answer 1

int n = 0; This initialization is not used. It could simply be int n;, or could be int n = !(num & 1); inside the loop, to restrict the scope of n.

---

This loop:

int k = 0;
while (num) {
 ...
 k++;
}

could be written as:

for(int k = 0; num; k++) {
 ...
}

---

Since you are doing bit manipulation, instead of using addition, you should probably use a "binary or" operation to merge the bit into your accumulator:

 b = b | (n << k);

or simply:

 b |= n << k;

---

Bug

You are not inverting the most significant zero bits. Assuming an 8-bit word size, the binary compliment of 9 (0b00001001) should be 0b11110110, not 0b00000110. And the compliment of that should return to the original number (0b00001001), but instead yields 0b00000001.

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And, as mentioned by @Martin R, you could simply return ~num;