TITLE: null and pointer representations

(Newsgroups: comp.std.c++, 7 Dec 98)


KELLNER: "Alfred Kellner" <alfkellner@magnet.at>

>> T*t=0; Has been meant as an address designating the very first
>> element of the lagest array of the smallest type that can exist in a given
>> address space - not as claim *that* address *is* the NULL pointer, in
>> fact, it cannot be (..is distinguishible from every other value of
>> pointer..).

CLAMAGE: Steve Clamage <stephen.clamage@sun.com>

> In fact it MUST be a null pointer. A language requirement is that
> a literal zero assigned to a pointer type yields a null pointer
> of that type, distinguishable from a pointer to any object (or
> function) of that type.
>
> For example, a simple implementation on a system with a linear
> address space is for address zero to be reserved, and not used
> as the address of any entity in the program.
>
> An implemenation can choose to use something else for a null
> pointer. In that case, assigning a literal zero to a pointer
> object does not result in all-bits-zero, but in some other
> value that is never used for any entity in the program.

KELLNER:

>> However, the question is still : Is an address a number ?

???:

>>> Furthermore, your use of the term ordinal implies that arbitrary pairs
>>> of addresses can be ordered.

KELLNER:

>> Agreed, that's implied in pointer arithmetic ++ptr gets the next ptr in
>> sequence, and ptr+n the (1+n)-th.

CLAMAGE: 

> No, that operation is only well-defined for objects within the
> same array. The operation result is not defined once you go outside
> the bounds of the array. (The address of array+n for an array of
> n elements is defined.)
>
> Arbitrary pairs of of addresses need not have any such relationship.

???:

>>> There are implementations where that doesn't work either.

KELLNER:

>>*This* comes as a very surprise to me. Pointer arithmetic doesn't work ?

CLAMAGE:

> It's only defined when the results remain within the same complete
> object. Example:
>   char A[100];
>   char *P = A + 101; // 2 past the end of A
> The addition of A and 101 is not guaranteed to produce a result;
> if it does, the assignment to P is not guaranteed to succeed;
> if it does, no operation on P except a new assignment has any
> defined result. That is, you are not guaranteed even to be able
> to inspect the value of the pointer P.