Why don't I get an integer overflow when adding two chars?

Neither C++ not C perform arithmetical computations withing "smaller" integer types like, char and short. These types almost always get promoted to int before any further computations begin. So, your expression is really evaluated as

unsigned char c = ((int) a + (int) b) / 2;

P.S. On some exotic platform where the range of int does not cover the range of unsigned char, the type unsigned int will be used as target type for promotion.

No, this is not an error.

The compiler always calculates at minimum of integer precision, the result will be converted back to unsigned char on assignment only.

This is in the standard.