# Change sign using bitwise operators

How to change the sign of int using bitwise operators? Obviously we can use x*=-1 or x/=-1. Is there any fastest way of doing this?

I did a small test as below. Just for curiosity...

public class ChangeSign { public static void main(String[] args) { int x = 198347; int LOOP = 1000000; int y; long start = System.nanoTime(); for (int i = 0; i < LOOP; i++) { y = (~x) + 1; } long mid1 = System.nanoTime(); for (int i = 0; i < LOOP; i++) { y = -x; } long mid2 = System.nanoTime(); for (int i = 0; i < LOOP; i++) { y = x * -1; } long mid3 = System.nanoTime(); for (int i = 0; i < LOOP; i++) { y = x / -1; } long end = System.nanoTime(); System.out.println(mid1 - start); System.out.println(mid2 - mid1); System.out.println(mid3 - mid2); System.out.println(end - mid3); } }

Output is almost similar to :

2200211 835772 1255797 4651923

## Answers

The speed difference between non-floating point (e.g. int math) addition/multiplication and bitwise operations is less than negligible on almost all machines.

There is no general way to turn an n-bit signed integer into its negative equivalent using only bitwise operations, as the negation operation looks like x = (~x) + 1, which requires one addition. However, assuming the signed integer is 32 bit you can probably write a bitwise equation to do this calculation. **Note**: do not do this.

The most common, readable way to negate a number is x = -x.

Java uses Complement Two representation. In order to change a sign, it means you must do a bitwise negation (it would be equivalent to xor with FFFF) and add 1.

x = ~x + 1;

I am almost sure that -x is, if anything, faster than that.

Solution using high level language

Questions like these are popular in interviews and competitive programming world .

I landed here researching more solution for negation of a number without using - or + operator .

For this :

- complement a number using ~ operator
Then add 1 to the number obtained in step 1 using Half adder logic :

int addNumbers(int x, int y) { if(y==0) return x; // carry is 0 return addNumbers(x^y,(x&y)<<1); }

Here x^y performs addition of bits and x&y handles carry operation