On 2023-12-12 at 09:11, Yongwei Wu via Std-Discussion wrote:
> On Tue, 12 Dec 2023 at 14:14, Daniel Krügler <daniel.kruegler_at_[hidden]> wrote:
>>
>> Am Di., 12. Dez. 2023 um 05:42 Uhr schrieb Yongwei Wu via
>> Std-Discussion <std-discussion_at_[hidden]>:
>>>
>>> In the case of erase in the following form:
>>>
>>> constexpr iterator erase(const_iterator);
>>>
>>> basic_string does not say anything about iterator and reference
>>> invalidation, while vector specifically says erase "invalidates
>>> iterators and references at or after the point of the erase".
>>
>> That's not quite right. We have [string.require] p4, which says:
>>
>> 4 References, pointers, and iterators referring to the elements of a
>> basic_string sequence may be invalidated
>> by the following uses of that basic_string object:
>> (4.1) — Passing as an argument to any standard library function taking
>> a reference to non-const basic_string
>> as an argument.
>> (4.2) — Calling non-const member functions, except operator[], at,
>> data, front, back, begin, rbegin, end,
>> and rend.
>>
>> So, bullet (4.2) applies for erase, which is a non-const member
>> function that is not excluded by the listed functions.
>
> Thank you.
>
> I would argue this is quite weak and vague. vector::erase specifically
> says it "invalidates iterators and references at or after the point of
> the erase". That implies iterators pointing to elements before the
> point of erase remain valid. That would be stronger than the point you
> mention here. It does not make sense to me that string provides a
> weaker guarantee than vector.
>
> And the main part of the questions is why (under what circumstance).
> Why should an iterator pointing to the point of erase be invalidated
> for the contiguous sequence containers? The memory layout is fixed,
> after all.
>

std::string allows the Short String Optimization (which vector doesn't),
so the memory layout is not guaranteed to be fixed.