Inconsistent behaviour of implicit conversion between unsigned and bigger signed types

Question

Consider following example:

#include 

int main(void)
{
    unsigned char a  = 15; /* one byte */
    unsigned short b = 15; /* two bytes */
          


        

Answer 1

Negatives are tricky. Especially when it comes to unsigned values. If you look at the c-documentation, you'll notice that (contrary to what you'd expect) unsigned chars and shorts are promoted to signed ints for computing, while an unsigned int will be computed as an unsigned int.

When you compute the -c, the c is treated as an int, it becomes -15, then is stored in x, (which still believes it is an UNSIGNED int) and is stored as such.

For clarification - No ACTUAL promotion is done when "negativeing" an unsigned. When you assign a negative to any type of int (or take a negative) the 2's compliment of the number is instead used. Since the only practical difference between unsigned and signed values is that the MSB acts as a sign flag, it is taken as a very large positive number instead of a negative one.