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

int is special. Everything smaller than int gets promoted to int in arithmetic operations.

Thus -a and -b are applications of unary minus to int values of 15, which just work and produce -15. This value is then converted to long.

-c is different. c is not promoted to an int as it is not smaller than int. The result of unary minus applied to an unsigned int value of k is again an unsigned int, computed as 2^N-k (N is the number of bits).

Now this unsigned int value is converted to long normally.