what is the representation of a signed number in unsigned form ?

[radha gogia]

main()
 
{
 
unsigned i=3;
 
if(i>-1)
 
    printf("Hell");
 
else
 
   printf("Heaven");
 
}

why is the output Heaven ,plz explain the logic for this .

[weaknessforcats]

Usually, the leftmost bit of the signed int is the sign bit. In unsigned form, this bit becomes part of the value.

Also, negative signed ints are usually in in 2's complement form so they would appear as very large negative numbers.

[Luuk]

main() {
 
unsigned i=3;
 
printf("i= %u\n", i);
printf("i= %u\n", -1);
 
if( i > -1 )
        printf("Hell");
    else
        printf("Heaven");
}
 

check: this "In i > -1, the -1 is converted to unsigned int, resulting in the value UINT_MAX. i is never bigger than that value, so the loop body never executes."

[donbock]

Line 7 of the OP compares i to -1. i is an unsigned int, -1 is a signed int. They are both flavors of int, so they have the same rank. One of the rules for the Usual Arithmetic Conversions is "if the operand that has unsigned integer type has rank greater or equal to the rank of the type of the other operand, then the operand with signed integer type is converted to the type of the operand with unsigned integer type." Thus, -1 gets converted from int to unsigned int before the comparison takes place.

C99 mandates that C only permit three possible encoding techniques for integer values:

• 2's-complement
• 1's-complement
• Sign-magnitude

If we assume ints have 16 bits, then

 Technique      (unsigned int)3   (int)(-1) (unsigned int)(-1)
2's-complement     0x0003         0xFFFF      65535  
1's-complement     0x0003         0xFFFE      65534
sign-magnitude     0x0003         0x8001      32769
 

In every case, casting -1 to an unsigned int yields a value greater than 3.