Can anyone explain type promotion of this program and why output is same

Question

Why do these two numbers compare as equal and how is the comparison between x and y performed?

#include <stdio.h>
int  main() 
{ 
   unsigned int x = -1; 
   int y = ~0; //type promotion from int to unsigned int
   if (x == y) 
      printf("same"); 
   else
      printf("not same"); 
   return 0; 
}

Answer 1

The ~ operator in C++ (and other C-like languages like C and Java) performs a bitwise NOT operation - all the 1 bits in the operand are set to 0 and all the 0 bits in the operand are set to 1. In other words, it creates the complement of the original number. For more details see here.

The bitwise NOT operator has an interesting property that when applied on numbers represented by two's complement, it changes the number's sign and then subtracts one (as you can see in the above example).

If you check what the ~-operator does to y like this:

#include <stdio.h>

int  main() 
{ 
   unsigned int x = -1; 
   printf("%i\n",x); // treat data as if singed integer
   printf("%u\n",x); // treat data is if unsigned integer

   int y = ~0; //type promotion from int to unsigned int
   printf("%i\n",y);
   if (x == y) 
      printf("same"); 
   else
      printf("not same"); 
   return 0; 
}

You can see:

Success time: 0 memory: 16064 signal:0

-1 same

Try out here.

Why is this so?

The bit representation for -1 in integer is (see two's complement for details)

11111111 11111111 11111111 11111111 

If you read this in unsigned integer

11111111 11111111 11111111 11111111

is the largest possible unsigned int. -1 in unsigned int becomes the maximal unsigned int number due to wrap around. Therfore unsigned -1 and int ~0 are the same.