Why does this code for incrementing an uint8_t include `& 0xFF`?

This kind of code is common when you want to avoid problems with implicit type promotions, or when you simply wish to demonstrate that you have considered implicit promotions when you wrote the code, which is good programming practice.

uint8_t is a small integer type and therefore always promoted to int whenever you use it in an expression. The result of (value + 1) is always int.

Without the masking, some compilers give warnings such as "attempting to store int in uint8_t". I've encountered such warnings on several compilers. Theoretically int & 0xFF is still an int, but since it cannot have a value larger than 0xFF, the compiler is likely able to optimize the type down to uint8_t and the warning will go away.

Alternatively you could write value = (uint8_t)(value + 1u); which has the same meaning (but is a MISRA-C compatible version of the code).

My guess is that this code was intended to work correctly even if value is not a 1-byte (8-bit) type. The bitmask 0xFF makes sure that only the last byte of the value is kept.