How does unsigned addition handle overflow?

It wraps: the result is (x + y) mod 2^w — any carry beyond bit w−1 is simply dropped.

* Unsigned values live on a clock of 0..2^w−1: adding 12 + 7 runs off the top and wraps around to 19 mod 16 = 3. *

If sum ≥ 2^w, the result wraps: result = (x + y) mod 2^w

Example (4-bit, max=15):

12 + 7 = 19, but 19 mod 16 = 3
Binary: 1100 + 0111 = 10011 → keeps only 0011

Detecting overflow: If result < either operand, overflow occurred.

// Check for unsigned overflow
unsigned sum = x + y;
if (sum < x) { /* overflow! */ }

Go deeper:

• Integer overflow (Wikipedia) — wraparound, its security consequences, and why unsigned is modular.