BItwise Operator, how does Left shift and right shift works

Question

public class Solution {
    public int GetSum(int a, int b) {
        while(b != 0) {
            int carry = a & b;
            a = a ^ b;
            b = carry << 1;
        }
        return a;
    }
}

So the question is to add two numbers without using arithmetic operators, I found a solution in leetcode discuss section but I am not sure how the carry variable works

Answer 1

public class Solution {
    public int GetSum(int a, int b) {
       // Iterate till there is no carry 
        while(b != 0) {
            // carry now contains common 
            // set bits of a and b 
            int carry = a & b;
             // Sum of bits of a and  
            // b where at least one  
            // of the bits is not set 
            a = a ^ b;
            // Carry is shifted by  
            // one so that adding it  
            // to a gives the required sum 
            b = carry << 1;
        }
        return a;
    }
}

example

when you call sum function like this then

GetSum(60, 13)
/* 60 = 0011 1100 */  
/* 13 = 0000 1101 */

while(b != 0) (its while loop and it will loop until b becomes 0)

int carry = a & b;; // a=60 & 13 

will become 12 /* 12 = 0000 1100 */

a = a ^ b;

will become 49 /* 49 = 0011 0001 */

b = carry << 1;

will become 24 because of Binary Left Shift Operator(<<). The left operands value is moved left by the number of bits specified by the right operand and it will loop and calculate continue like i explained