C usual arithmetic conversions

Question

I was reading in the C99 standard about the usual arithmetic conversions.

If both operands have the same type, then no further conversion is needed.

Otherwise, if both operands have signed integer types or both have unsigned integer types, the operand with the type of lesser integer conversion rank is converted to the type of the operand with greater rank.

Otherwise, 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.

Otherwise, if the type of the operand with signed integer type can represent all of the values of the type of the operand with unsigned integer type, then the operand with unsigned integer type is converted to the type of the operand with signed integer type.

Otherwise, both operands are converted to the unsigned integer type corresponding to the type of the operand with signed integer type.

So let's say I have the following code:

#include <stdio.h>

int main()
{
    unsigned int a = 10;
    signed int b = -5;

    printf("%d\n", a + b); /* 5 */
    printf("%u\n", a + b); /* 5 */
    return 0;
}

I thought the bolded paragraph applies (since unsigned int and signed int have the same rank. Why isn't b converted to unsigned ? Or perhaps it is converted to unsigned but there is something I don't understand ?

Thank you for your time :-)

Answer 1

Indeed b is converted to unsigned. However what you observed is that b converted to unsigned and then added to 10 gives as value 5.

On x86 32bit this is what happens

1. b, coverted to unsigned, becomes 4294967291 (i.e. 2**32 - 5)
2. adding 10 becomes 5 because of wrap-around at 2**32 (2**32 - 5 + 10 = 2**32 + 5 = 5)

Answer 2

0x0000000a plus 0xfffffffb will always be 0x00000005 regardless of whether you are dealing with signed or unsigned types, as long as only 32 bits are used.