I have been confused about bit manipulation for a very long time. So I decide to do a summary about it here.

"&" AND operation, for example, 2 (0010) & 7 (0111) => 2 (0010)

"^" XOR operation, for example, 2 (0010) ^ 7 (0111) => 5 (0101)

"~" NOT operation, for example, ~2(0010) => -3 (1101) what??? Don't get frustrated here. It's called two's complement.

1111 is -1, in two's complement

1110 is -2, which is ~2 + 1, ~0010 => 1101, 1101 + 1 = 1110 => 2

1101 is -3, which is ~3 + 1

so if you want to get a negative number, you can simply do ~x + 1

Reference:

https://en.wikipedia.org/wiki/Two%27s_complement

https://www.cs.cornell.edu/~tomf/notes/cps104/twoscomp.html

For this, problem, for example, we have a = 1, b = 3,

In bit representation, a = 0001, b = 0011,

First, we can use "and"("&") operation between a and b to find a carry.

carry = a & b, then carry = 0001

Second, we can use "xor" ("^") operation between a and b to find the different bit, and assign it to a,

Then, we shift carry one position left and assign it to b, b = 0010.

Iterate until there is no carry (or b == 0)

```
// Iterative
public int getSum(int a, int b) {
if (a == 0) return b;
if (b == 0) return a;
while (b != 0) {
int carry = a & b;
a = a ^ b;
b = carry << 1;
}
return a;
}
// Iterative
public int getSubtract(int a, int b) {
while (b != 0) {
int borrow = (~a) & b;
a = a ^ b;
b = borrow << 1;
}
return a;
}
// Recursive
public int getSum(int a, int b) {
return (b == 0) ? a : getSum(a ^ b, (a & b) << 1);
}
// Recursive
public int getSubtract(int a, int b) {
return (b == 0) ? a : getSubtract(a ^ b, (~a & b) << 1);
}
// Get negative number
public int negate(int x) {
return ~x + 1;
}
```