Sum of Two Integers without the + operator.

Problem. Compute a + b without using the + or - operators. Use bit operations only.

The insight — split into sum-without-carry and carry

Binary addition decomposes into two parts:

• Sum without carry: a ^ b — XOR adds bit-by-bit but ignores carries (1 + 1 = 0 with carry lost).
• Carry: (a & b) << 1 — a carry is generated exactly where both bits are 1, and it propagates to the next-higher position (hence the left shift).

Add the partial sum to the carry, repeating until the carry becomes zero:

int getSum(int a, int b) {
    while (b != 0) {
        int carry = (a & b) << 1;   // where both bits are 1, carry left
        a = a ^ b;                  // add without carrying
        b = carry;                  // fold the carry back in and repeat
    }
    return a;
}

When b (the carry) reaches 0, a holds the full sum.

Walkthrough — 3 + 5

a = 011 (3), b = 101 (5)
carry = (011 & 101) << 1 = 001 << 1 = 010 ;  a = 011 ^ 101 = 110 ;  b = 010
carry = (110 & 010) << 1 = 010 << 1 = 100 ;  a = 110 ^ 010 = 100 ;  b = 100
carry = (100 & 100) << 1 = 100 << 1 = 1000;  a = 100 ^ 100 = 000 ;  b = 1000
carry = (000 & 1000)<< 1 = 0          ;       a = 000 ^ 1000 = 1000;  b = 0
-> a = 1000 = 8

Subtraction and negatives

To subtract, add the two's complement: a - b == getSum(a, getSum(~b, 1)). The loop already handles negative operands correctly because Java's int is two's complement and << / ^ / & operate on the raw bit patterns — no special case needed. The carry is guaranteed to terminate within 32 iterations.