That is true but it's not the reason... INT_MAX + 1 must be 0 for unsigned int because the C standard defines wrap-around behavior for unsigned data types. Saturated data types are only defined in Embedded C. E.g.: Sat Fract
That is the entire reason. Because the representation of signed numbers could vary from implementation to implementation, making INT_MAX + 1 (or any signed integer overflow) undefined allows for more efficient generated code. If it were defined in some way, the generated code would need checks for overflow wherever it was possible, which would lead to bigger and slower code.
INT_MAX + 1 is never 0 for unsigned int. (unsigned)INT_MAX + 1 is equal to INT_MAX + 1. You're thinking of UINT_MAX + 1, which is always 0.
It's not true that (unsigned)INT_MAX + 1 is never 0 - it's allowed for UINT_MAX == INT_MAX (leaving the sign bit unused in the unsigned representation). I've never seen such an implementation, though.
