integral overflow and integral conversions

[Jason Merrill]

On 11/07/2010 11:11 PM, Ian Lance Taylor wrote:
> C99 6.2.5 paragraph 9 says "A computation involving unsigned operands
> can never overflow, because a result that cannot be represented by the
> resulting unsigned integer type is reduced modulo the number that is one
> greater than the largest value that can be represented by the resulting
> type."
>
> In the C++0x draft I have, 3.9.1 paragraph 4 says "Unsigned integers,
> declared unsigned, shall obey the laws of arithmetic modulo 2**N where n
> is number of bits in the value representation of that particular size of
> integer."
>
> So I don't think unsigned arithmetic can overflow in C/C++.

Ah, so I was missing something, thanks!

Jason