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Re: What is the meaning of `new T[0]`?

From: language.lawyer_at <language.lawyer_at_[hidden]>
Date: Sat, 11 Sep 2021 00:41:29 +0500

On 11/09/2021 00:11, Brian Bi via Std-Discussion wrote:
> It's explicitly allowed to create an array with 0 elements using a
> new-expression, but the standard doesn't seem to explain the meaning of the
> returned pointer.
> As I see it, the most literal reading of the standard is the following: the
> new-expression creates an array object with 0 elements. Because an object
> can't have size 0, this empty array will have some nonzero
> implementation-defined size. The pointer that's returned by the
> new-expression points to this array. Since a pointer value is exactly one
> of {null, pointer to object/function, pointer past the end, invalid}, and
> this pointer does point to the array, it is not a past-the-end pointer.
> Perhaps it was intended for this pointer to be in the past-the-end
> category, but the current wording doesn't seem clear enough.

Received on 2021-09-10 14:41:35