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From: James Kuyper <jameskuyper@alumni.caltech.edu>
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Date: Mon, 8 Jul 2024 00:28:53 -0400
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On 7/7/24 20:02, Kaz Kylheku wrote:
....
> Ritchie's B language had arrays which contained a pointer to their
> first element. Via a hack, it was possible to relocate an array.
> 
> In C, such a thing is not simply not required; it is ruled out
> by the detailed semantic description of arrays.
> 
> The entire representation of an array of size N elements of type
> T is contained in the memory block that is sizeo(T)*N bytes wide.
> 
> If you copy that block, you have a fully functional copy of the array.
> No extra pointer needs to be set up with the correct value.

An implementation which took the following code:

int array1[5], array2[5];
memcpy(array2, array1, sizeof array1);

and translated it into machine code that was the equivalent of

int array1[5], array2[5];
int *_p1 = &array1[0], *_p2 = &array2[0];
memcpy(_p2, _p1, sizeof array1);

would not violate any of the requirements you mention. The key point is
that when you copy the contents of an array to a new location, you
wouldn't want to copy the implicit pointer - it would point at the wrong
location. And if the destination is itself declared as an array, it
would already have an implicit pointer that pointed at the correct location.

I see no point in having implicit pointers, but I don't believe that
they are prohibited.

> Furthermore, to dynamically allocate an array, you need only
> provide sizeof(T)*N bytes of storage, and not a bit more.
> 
> There is simply nowhere in the representation of an array where
> a pointer could hide that is part of the representation.

Allocated memory is already accessed through a pointer; there would be
no corresponding need to create an implicit one when there's already an
explicit one.