What is a ^ b and (a & b) << 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 ⊕ bconst carry=(a&b)<<1;       // carry = 2*(a and b), since we carry to the next bitif(!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

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

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

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

01010110------0011

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

01010110-------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,

00111000-------1011

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

00111000-------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.