Here is a conjecture:
For expression a op b where a and b are of the same unsigned integral type U, and op is one of the compound assignment operators (+=,-=,*=,/=,%=,&=,|=,^=,<<=,>>=), the result is computed directly in the value domain of U using modular arithmetic, as if no integral promotions or usual arithmetic conversions, etc. are performed at all.
Is this true? What about signed integral types?
To clarify: By definition integral promotions and usual arithmetic conversions do apply here. I'm asking if the result is the same as not applying them.
I'm looking for an answer in C++, but if you can point out the difference with C, it would also be nice.
Counter example:
int has width 31 plus one bit for sign, unsigned short has width 16. With a and b of type unsigned short, after integral promotions, the operation is performed in int.
If a and b have value 2^16 - 1, then the mathematical exact result of a * b in the natural numbers would be 2^32 - 2^17 + 1. This is larger than 2^31 - 1 and therefore cannot be represented by int.
Arithmetic overflow in signed integral types results in undefined behavior. Therefore a *= b has undefined behavior. It would not have this undefined behavior if unsigned arithmetic modulo 2^width(unsigned short) was used.
(Applies to all C and C++ versions.)