What is a ^ b and (a & b) << 1?

Question

I was doing this question in leetcode.

Request:

Calculate the sum of two integers a and b, but you are not allowed to use the operator + and -.

I can't understand the solution it gave

Could someone explain how this getSum function works?

Here is the answer in JS:

var getSum=function(a,b) {
    const Sum = a^b; //I can't understand how those two line's code can
    const carry = (a & b) << 1; //get the sum
        if(!carry) {
            return Sum
        }
    return getSum(Sum,carry);
};
console.log(getSum(5,1));

Answer 1

It's basically replicating the half-adder

Adding 2 bits A and B produces 2 outputs: a sum and a carry bit like below

╔═══════╤═════════════╗
║ Input │   Output    ║
╠═══╤═══╪═══════╤═════╣
║ A │ B │ carry │ sum ║
╟───┼───┼───────┼─────╢
║ 0 │ 0 │   0   │  0  ║
╟───┼───┼───────┼─────╢
║ 1 │ 0 │   0   │  1  ║
╟───┼───┼───────┼─────╢
║ 0 │ 1 │   0   │  1  ║
╟───┼───┼───────┼─────╢
║ 1 │ 1 │   1   │  0  ║
╚═══╧═══╧═══════╧═════╝

From the table we get the logic for the outputs: carry = A and B, sum = A xor B

XOR is also called a carry-less add operator, and represented by ⊕ with the + symbol inside

So the snippet above is working like this

const Sum=a^b;              // sum = a xor b = a ⊕ b
const carry=(a&b)<<1;       // carry = 2*(a and b), since we carry to the next bit
if(!carry){
    return Sum;             // no carry, so sum + carry = sum
}
return getSum(Sum,carry);   // a + b = sum + carry

So a^b adds each bit in a and b simultaneously, leaving the non-carry sum of a and b in Sum. Then we have to add carry to the carry-less sum to get the final result, since we have only a half-adder instead of a full-adder which does a + b = a ⊕ b + carry

See also

• Adding two numbers without + operator (Clarification)
• What is the best way to add two numbers without using the + operator?
• adds two numbers without using + or any arithmetic operators
• Adding two numbers without using the addition operator

Answer 2

Let's learn by example. Imagine that a = 3 and b = 5

In binary notation they are a = 0011 and b = 0101

XOR: a^b is XOR operator. When compare two bits it returns 0 if they are same and 1 if they are different. 01^10 => 11

So when we're doing a^b result will be 0110.

AND + SHIFT

a&b performs logical AND operation. It returns 1 only when a = b = 1.

In our case the result is 0001

<< shifts it(adds 0 on the right side) and result became 0010 which sets carry variable true. (only 0000 will be false).

Next iterations:

Everything repeats but now a = 0110 and b = 0010 (Sum and carry from last execution)

Now a^b = 0100 and (a&b)<<1 = 0100

Repeating again.

Now a^b = 0000 and (a&b)<<1 = 1000

And again.

Now a^b = 1000 and (a&b)<<1 = 0000. Now carry is finally false. And we're returning 1000 which is decimal 8.

Everything worked fine since 3+5=8

Answer 3

 int result = p ^ q; // XOR Operator, + without carry 0+0=0, 0+1=1+0=1, 1+1=0
int carry = (p & q) << 1; // Left Shift, 1+1=2
if (carry != 0) {
    return getSum(result, carry);
}
return result;

Start By p=5,q=6. Then the XOR would be,

0101
0110
------
0011

So, XORing results in (0011) which is actually 3 in decimal. Then ANDing p and q we get,

0101
0110
-------
0100

We get 4 (100 in binary) by ANDing 5 & 6, now if we left shift this value by 1, we get

 0100<<1=1000

So we get 8 (1000 in binary) after first recursion.As the result (carry variable) isnt zero, lets recursion again by xor value and carry value.

getSum(3, 8);

So, doing the first XORing we get,

0011
1000
-------
1011

The XORing this time yielded in 11 (1011 binary),so we perform the AND now,

0011
1000
-------
0000

We get all ZERO for ANDing 3 and 8, so this time the left shift operator also results in ZERO, as we have no 1 here which may give us a value by left shifing zeroes. As the carry variable is now Zero, we come to the end of recursion and the XORed value will be the Sum, which is 11 (1011 in Binary).

Hope you get the working of the procedure. You can learn more by learning bitwise operation, as its the way the machine do the arithmatic operations.

Answer 4

^ is XOR, a bitwise operation. On a single bit, the rules are 0 ^ 0 = 0, 0 ^ 1 = 1, 1 ^ 0 = 0, and 1 ^ 1 = 0, and you simply extend perform it on corresponding bits when dealing with multi-bit values. The name is short for "exclusive or", and comes from the fact that A ^ B is 1 if and only if either A or B is 1, not both. But, it's more interesting to talk about its other name, ⊕. ⊕ is + but slightly different. You'll notice that the rules for ⊕ are similar to the rules for addition: 0 + 0 = 0, 0 + 1 = 1, 1 + 0 = 1, and 1 + 1 = 10. ⊕ is +, except 1 ⊕ 1 = 0; that is, ⊕ is +, except without carrying. This holds for multiple bits: 011 + 001 = 100, because you carry a 1 out of the ones place into the twos place, and then carry a 1 again into the fours place. Then, 011 ⊕ 001 = 010, because you just don't carry.

Now, when doing real addition, when do you carry? In binary, the answer is very simple: you carry a 1 into the next place when there are two 1s in a given place. This is easily understood as a bitwise AND, &. 1 & 1 = 1, and 0 otherwise. For 011 + 001, addition without carrying gives 011 ⊕ 001 = 010, and we can tell we need to carry a 1 out of the ones place because 011 & 001 = 001. The shifting in (a & b) << 1 turns a number "where do I need to carry from?" into "where do I need to add carries?": (011 & 001) << 1 = 010; I need to add a carry bit in the twos place.

So, in getSum, we want to know a + b. We compute the addition without carrying with a ^ b, and we find where we need to add carry bits with (a & b) << 1. Now, we just need to add those two together. Well, we already have a function for adding numbers together; it's called getSum. So, we basically just write function getSum(a, b) { return getSum(a ^ b, (a & b) << 1); }, except we make sure to short-circuit if there is nothing to carry, saving us from infinite recursion.