How does the last integer promotion rule ever get applied in C?
- by SiegeX
6.3.1.8p1: Otherwise, the integer promotions are
performed on both operands. Then the following rules are applied to the promoted operands: If both operands have the same
type, then no further conversion is needed. Otherwise, if both operands have signed integer types or both have unsigned integer
types, the operand with the type of lesser integer conversion rank is converted to the type of the operand with greater rank.
Otherwise, if the operand that has unsigned integer type has rank greater or equal to the rank of the type of the other operand,
then the operand with signed integer type is converted to the type of the operand with unsigned integer type. Otherwise,
if the type of the operand with signed integer type can represent all of the values of the type of the operand with unsigned
integer type, then the operand with unsigned integer type is converted to the type of the operand with signed integer type.
Otherwise, both operands are converted to the unsigned integer type corresponding to the type of the operand with signed
integer type.
For the bolded rule to be applied it would seem to imply you need to have have an unsigned interger type who's rank is less than the signed integer type and the signed integer type cannot hold all the values of the unsigned integer type.
Is there a real world example of such a case or is this statement serving as a catch-all to cover all possible permutations?
