On 7/11/25 01:28, Brian Bi wrote:
>
>
> On Thu, Jul 10, 2025 at 5:57 PM Ell via Std-Discussion <std-
> discussion_at_[hidden] <mailto:std-discussion_at_[hidden]>> wrote:
>
> If an array provides storage for an object, does destroying the object
> "invalidate" the contents of the array?
>
> void f () {
> unsigned char a[1];
>
> using T = unsigned char;
>
> T* p = new (a) T (123);
> unsigned char x = a[0];
> p->~T ();
>
> unsigned char y = a[0];
>
> assert (x == y);
> }
>
>
> In this case, we have to answer the question of whether the newly
> created object is an element of the array, or whether the original
> element is still there and the array is providing storage for the new
> object.
>
> [intro.object]/2 <http://eel.is/c++draft/intro.object#2> states that the
> new object in this particular case is a subobject of the array `a`.
> Presumably an array cannot have two elements with index 0, so we can
> infer that the original `a[0]`'s lifetime ends as in [basic.life]/2.5
> <http://eel.is/c++draft/basic.life#2.5>. The array doesn't provide
> storage for the new `unsigned char` object. When the new object is later
> destroyed, `a[0]` is out-of-lifetime and attempting to access its value
> results in UB.
>
> If you had created an object of some other type, say, `char`, in the
> storage of `a`, the standard is unclear as to what happens to the value
> of `a[0]` in that case (either during or after the `char` object's
> lifetime).

Huh, I didn't even consider this distinction. I didn't specifically mean
for the types to be the same. Though, the way I'm reading
[intro.object], paragraphs 2 and 3 aren't mutually exclusive, and the
code fits both (whether or not this makes sense).