Negation error in 16 bit 2's complement code - c

Question

It's a code for finding 2's complement. When num is char, it works fine for 8 bit. But when it is unsigned int for 16-bits numbers. Code doesn't work.
Can you guide?
#include<stdio.h>
#define ISNEGATIVE(num) num & 0x80
#define TWO_COMP(num) (~num) + 1
int main()
{
unsigned int num;
num = 0xFFFF;
if(ISNEGATIVE(num))
printf("1's = %d, 2's = %d", ~num, TWO_COMP(num));
return 0;
}
Output :
1's = -65536, 2's = -65535

Answer 1

An unsigned int is a 32-bit variable on most computers. If you want to use a 16-bit value use uint16_t by preference, or unsigned short. You also need to use the appropriate half-word specifiers in your printf call:
#include <stdio.h>
#include <stdint.h>
#define ISNEGATIVE(num) num & 0x8000
#define TWO_COMP(num) (~num) + 1
int main()
{
/*unsigned short num;*/
uint16_t num;
num = 0xFFFF;
if(ISNEGATIVE(num))
printf("1's = %hd, 2's = %hd", ~num, TWO_COMP(num));
return 0;
}
This outputs
1's = 0, 2's = 1

Answer 2

The 2 primary issues are:
the width of the mask needs to match the width of the number being looked at:
this algorithm should be limited to testing signed types,
Width of char AKA int8_t is 8 bits, so 0x80 works fine.
Width of short int AKA int16_t is 16 bits so requires 0x8000
Width of int AKA int32_t is 32 bits so would require 0x80000000
and so on...
Because num in your code is 16 bits wide, the line:
#define ISNEGATIVE(num) num & 0x80
Should be:
#define ISNEGATIVE(num) num & 0x8000
Just as important, because the value stored in the MSB of an unsigned type does not make the value negative, the algorithm should only be used to evaluate signed types.
unsigned int num;
Should be:
int num;