Why is the format specifier for uint8_t and uint16_t the same (%u)? [duplicate]

Answer 1

Because %u stands for "unsigned", it well may be uint64_t and is architecture dependent. According to man 3 printf, you may want to use length modifier to get sought behaviour, i.e. %hu (uint16_t) and %hhu (uint8_t).

Answer 2

printf() is a variadic function. Its optional arguments( and only those ) get promoted according to default argument promotions( 6.5.2.2. p6 ).

Since you are asking for integers, integer promotions are applied in this case, and types you mention get promoted to int. ( and not unsigned int because C )

If you use "%u" in printf(), and pass it an uint16_t variable, then the function converts that to an int, then to an unsigned int( because you asked for it with %u ) and then prints it.