The particular example is not something i reproduced here since it was
prop’… suffice to say they aren’t identical types ;)

I also understand there is clever way to achieve that result, I mentioned
the intent is to offer syntactic sugar & nothing else.

I’m not reading there is a proposal in flight that discuss this however, so
I’ll take it there was little interest so far.

Thanks all.
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On Wed, Apr 30, 2025 at 4:41 Zhihao Yuan via Std-Proposals <
std-proposals_at_[hidden]> wrote:

> On Tuesday, April 29th, 2025 at 12:17 PM, Arthur O'Dwyer <
> arthur.j.odwyer_at_[hidden]> wrote:
>
> On Tue, Apr 29, 2025 at 3:06 PM Zhihao Yuan via Std-Proposals <
> std-proposals_at_[hidden]> wrote:
>
>>
>> To push this idea further, you can do this in C++26
>>
>> template <class It1, class It2, class It3 = decltype(std::next(std::declval<It1>()))>
>> struct first_last_next
>> {
>> It1 first;
>> It2 last;
>> It3 next = std::next(first);
>>
>> explicit constexpr operator bool() const noexcept
>> {
>> return first != last and first != next;
>> }
>> };
>>
>> if (auto [it, _, itt] = first_last_next{v1.begin(), v1.end()}) {
>>
>> (* just a contrived snippet to show the possibility;
>> std::next doesn't return an iterator of a different type. *)
>>
>
> Wellll... IIUC, the conversion-to-boolean *will* happen before the
> destructuring access to `__temp.next` (good!), but the *initialization*
> of `__temp`, including `__temp.next`, will still happen before *that*. So
> you end up evaluating `std::next(first)` even when `first == last`, and
> thus you have UB for e.g. vector iterators.
> IIUC, the above is essentially equivalent to:
>
> auto first = v1.begin();
> auto last = v1.end();
> auto next = std::next(first); // oops, UB!
> if (first != last && first != next) {
> auto& it = first;
> auto& itt = next;
> [...]
>
>
> That is the top-of-the-thread's intent,
>
> - I want this
> if (auto it = vec.begin() /*and*/ auto itt = ++vec.begin(); it !=
> vec.end() && it != itt) {
>
>
> so I did not change that.
>
> To make this tricky code actually work, you'd have to make the call to
> `std::next` happen inside `get<2>(__temp)` rather than inside the
> initialization. Or, I guess, just make the initializer
> It3 next = (first == last) ? first : std::next(first);
>
> Right?
>
>
> I need a legitimate and complete example to
> know whether a transformation is equivalent 🤷
>
> --
> Zhihao Yuan, ID lichray
> The best way to predict the future is to invent it.
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