Hi
On 23-02-18 00:02, Selva Nair wrote:
> On Thu, Feb 22, 2018 at 5:37 PM, Selva Nair <selva.n...@gmail.com> wrote:
>>> +/** Return true if the addition of a and b would overflow. */
>>> +static inline bool
>>> +time_t_add_overflow(time_t a, time_t b) {
>>> + static_assert(((time_t) -1) < 0, "OpenVPN assumes time_t is signed");
>>> + static_assert(((time_t) .9) == 0, "OpenVPN assumes time_t is integer
>>> type");
>>> + static_assert(sizeof(time_t) == sizeof(long) || sizeof(time_t) ==
>>> sizeof(long long),
>>> + "OpenVPN assumes that time_t is of type long int or long long
>>> int");
>>> + static const time_t TIME_MAX = sizeof(time_t) == sizeof(long) ?
>>> + LONG_MAX : LLONG_MAX;
>>> + static const time_t TIME_MIN = sizeof(time_t) == sizeof(long) ?
>>> + LONG_MIN : LLONG_MIN;
>>> + return (a > 0 && b > TIME_MAX - a) || (a < 0 && b < TIME_MIN - a);
>>
>> Interesting. But I think this can be simplified much. Addition of
>> identically sized integers a and b overflows if and only if
>>
>> ((a > 0 && a + b < b) || (a < 0 && a + b > b))
>>
>> As overflow is possible only when both have same sign it can also be written
>> as
>>
>> ((a > 0 && a + b < a) || (a < 0 && a + b > a))
>>
>> So the TIME_MAX and TIME_MIN may be eliminated and that means no need
>> to check signed/unsigned or long/long-long.
>>
>> Am I missing something?
>
> Hm... replying to self: I suppose the worry is related to unsigned int
> arithmetic overflow being undefined behaviour in C. So potentially a
> compiler can treat those statements as always true if it wishes..
>
> Well, excuse the noise I caused then.
Yeah, at least, *signed* integer overflow is undefined, and time_t is
(usually) a signed type.
-Steffan
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