On Friday, 13 November 2015 at 09:09:33 UTC, Don wrote:
At the very least, we should change the terminology on that
page. The word "overflow" should not be used when referring to
both signed and unsigned types. On that page, it is describing
two very different phenomena, and gives the impression that it
was written by somebody who does not understand what they are
talking about.
The usage of the word "wraps" is sloppy.
That page should state something like:
For any unsigned integral type T, all arithmetic is performed
modulo (T.max + 1).
Thus, for example, uint.max + 1 == 0.
There is no reason to mention the highly misleading word
"overflow".
For a signed integral type T, T.max + 1 is not representable in
type T.
Then, we have a choice of either declaring it to be an error,
as C does; or stating that the low bits of the
infinitely-precise result will be interpreted as a two's
complement value. For example, T.max + 1 will be negative.
(Note that unlike the unsigned case, there is no simple
explanation of what happens).
Please let's be precise about this.
I don't understand what you think is so complicated about it?
It's just circular boundary conditions. Unsigned has the
boundaries at 0 and 2^n - 1, signed has them at -2^(n-1) and
2^(n-1) - 1.
Less straightforwardly, but if you like modular arithmetic:
After arithmetic operations f is applied
unsigned: f(v) = v mod 2^n - 1
signed: f(v) = ((v + 2^(n-1)) mod (2^n - 1)) - 2^(n-1)
