3 ms·

(u)int8_t have the same problems, including aliasing because they are just alias of (unsigned) char. Sometimes it's nice to have modular arithmetic mod 256, or

• wahern 7y ago
  If int8_t exists[1], then you know that char is 8 bits[2] and therefore know that char in char + char always promotes to int because int must have at least 16 value bits and a 16-bit int can represent any char value regardless of signedness. [1] int8_t is not required. [2] char is the fundamental unit of addressability. sizeof char always evaluates to 1, sizeof int8_t must be non-0, char must be at least 8 bits, and int8_t must be precisely 8 bits, therefore sizeof int8_t == sizeof char and CHAR_BIT == 8.