Well....
Signed integer arithmetic and unsigned integer arithmetic do not differ. At all.
The difference between an Int16 and a UInt16 is not in the 16 defined bits. It's in the infinite number of implicit bits representing place values above 2^15. Those bits are always 0 for the UInt16, but they're identical with the high bit of the Int16.
So it's not clear what it would mean for the result type to be "incorrect mathematically". Mathematically, an Int16 and a UInt16 are the same thing.
However, the path by which you promote the values does matter. I don't know what Julia does. But:
-10 (Int8) + 10 (UInt16)
-10 (Int16) + 10 (UInt16)
65526 (UInt16) + 10 (UInt16)
0 (UInt16)
is a different result from -10 (Int8) + 10 (UInt16)
246 (UInt8) + 10 (UInt16)
246 (UInt16) + 10 (UInt16)
256 (UInt16)
One of those should be the result Julia gives. I tend to hope it's the first one. That would correspond to a promotion strategy of "always expand the type to its full width before converting between signed and unsigned". Expansion (and shift, I guess) is the only operation for which the difference between signed and unsigned is relevant.> Would a better promotion be Int32 for both?
Probably not; that would imply that when you add two UInt16s together, you expect to get a UInt32 (or UInt17...) back.
Great point on the path dependence.
That depends on what you're hoping to do with the number you're looking for. If you wanted -5 as an Int16, the bit pattern would be 1111 1111 1111 1011 or 0xFFFB.
If you wanted 65531 as a UInt16, the bit pattern for that is 1111 1111 1111 1011 or 0xFFFB. You're getting the same result either way. And any values you compute from that result [that is, by arithmetic] are going to be unaffected by whether you labeled the result "Int16" or "UInt16", because that's just a label. If you labeled 0xFFFB a "Snerf", the arithmetic would still be the same.
There are only a very restricted set of places where you need to be explicit about whether you think of your variable as an Int or a UInt:
1. When you're widening it.
2. When you're doing a right shift.
3. When you're formatting it for display to a human.
Silently casting signed integers to unsigned is a terrible idea, and a recipe for bugs. And completely unnecessary to boot, because you could promote Int8 to Int16, and (Int16, UInt16) to (Int32, Int32).
unsigned x = (unsigned short)65535 * (unsigned short)65535;
(It overflows even though it's less than UINT_MAX, and worse it's UB.)