Janne Grunau <[email protected]> writes:

> On 2012-12-07 13:54:02 +0000, Måns Rullgård wrote:
>> The values compared here can be more than INT64_MAX apart.  Since the
>> difference is always positive, converting to uint64_t before subtracting
>> gives the correct result without overflows.
>
> s/(without)/\1 undefined/ it still relies on defined over/underflow
> behavior of unsigned types.

No, unsigned wraparound is not termed an overflow.

-- 
Måns Rullgård
[email protected]
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