Except that doing so does NOT avoid signed overflow. You're converting effectively w-1 bits of precision (plus a sign bit) to w bits of precision, doing the calculation, and then converting w bits of precision back to w-1 bits of precision -- that only works for a limited range of input values.

Using int32_t and uint32_t, try adding -2147483648 and -1. The arithmetic result should be -2147483649, but a program converting back and forth between signed and unsigned will produce 2147483647 -- wrong answer and wrong sign.