When adding two binary numbers by hand we keep the carry bits in mind and add it at the same time. But to do same thing in program we need a lot of checks. Recursive solution can be imagined as addition of carry and a^b (two inputs) until carry becomes 0.
Input : int x = 45, y = 45 Output : 90 Input : int x = 4, y = 78 Output : 82
Sum of two bits can be obtained by performing XOR (^) of the two bits. Carry bit can be obtained by performing AND (&) of two bits.
Above is simple Half Adder logic that can be used to add 2 single bits. We can extend this logic for integers. If x and y don’t have set bits at same position(s), then bitwise XOR (^) of x and y gives the sum of x and y. To incorporate common set bits also, bitwise AND (&) is used. Bitwise AND of x and y gives all carry bits. We calculate (x & y) << 1 and add it to x ^ y to get the required result.
One important observation is, if (x & y) becomes 0, then result is x ^ y.
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