On 17 August 2023 11:41:32 UTC, Ville Voutilainen via Std-Proposals <std-proposals_at_[hidden]> wrote:
>On Thu, 17 Aug 2023 at 14:25, Timur Doumler <cpp_at_[hidden]> wrote:
>
>> AFAIK it does however does *not* provide a guarantee that there is an actual array, that is, a T[N], under the hood anywhere. A hostile implementation could implement std::array<T, 2> for example as a struct { T first, second; };
>>
>>
>> No, it can't. That implementation wouldn't fulfill the requirements of
>> array::data(), which are that
>> [data(), data() + size())
>> is a valid range (and that range is denoted by pointer arithmetic, and
>> traversed by pointer arithmetic,
>> because data() and data()+size() are just pointers).
>>
>>
>> The implementation can fulfil this requirement by fiat – because the compiler makes it so.
>
>The pointer arithmetic is defined only for an array.
>
>>std::complex has this property, too – although it's specified differently (see [complex.numbers.general]/4), the effect is the same, you can use pointer arithmetic here, because the Standard says that you can.
>
>To do that, the standard would need to say something additional, like
>it does for std::complex, but since it doesn't, an array is there.
>
>>None of this guarantees that there is an actual T[N] anywhere under the hood, certainly not for std::array as far as I can see.
>
>Well, yeah - the actual problem is that nothing prevents the
>implementation from putting a T[N+IMPL_DEFINED_CONSTANT] there.
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Empty std::array does not have an underlying array ;)

Also, an std::array of one element can be implemented without an array.