Is arithmetic overflow equivalent to modulo operation?

Yes, the behavior of both of your examples is the same. See C99 6.2.5 §9 :

A computation involving unsigned operands can never overflow, because a result that cannot be represented by the resulting unsigned integer type is reduced modulo the number that is one greater than the largest value that can be represented by the resulting type.

No. There is nothing that guarantees that unsigned char has eight bits. Use uint8_t from <stdint.h>, and you'll be perfectly fine. This requires an implementation which supports stdint.h: any C99 compliant compiler does, but older compilers may not provide it.

Note: unsigned arithmetic never overflows, and behaves as "modulo 2^n". Signed arithmetic overflows with undefined behavior.