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.

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)