Re: Neatest way to get the end pointer?

vipps...@xxxxxxxxx wrote:
Keith Thompson <ks...@xxxxxxx> wrote:
vipps...@xxxxxxxxx writes:
"Tomás Ó hÉilidhe" <t...@xxxxxxxxxxx> wrote:
I commonly use pointers to iterate thru an array.
For example:
    int my_array[X];
[snip]
1) my_array is an int[X]
2) &my_array is an int(*)[X]
3) &my_array+1 is the address of the non-existant
array located after the current one.

And you invoke undefined behavior.

No, he doesn't.  &my_array+1 is a valid address,

Though it requires a conversion back to int * if used
as a pointer 'index' compared against an int * iterator.

just past the end of my_array.  Computing this address
is ok; attempting to dereference it would invoke UB.

I think that when the result of an expression is a non-
valid pointer, the behavior is undefined.

True, but one byte past the end of an array _is_ a valid
pointer.

Correct me if I am wrong,

Keith already has.

but I have also seen comments in GNU code like this:
--
/* ANSI C violation */
char * s = p - 1;

One byte beyond is fine (though you can't dereference it),
but there are no special rights for one (or more) byte(s)
_before_.

--
Where p points to the start of a string passed. 's' is
never dereferenced, ...

Doesn't matter. Implementations do not have to cope with
'one byte before'. An allocation can be made at the start
of a memory page. Merely calculating an address prior to
such a page may trap. In contrast, implementations _must_
cope with one byte beyond, which simply means that at
least one byte (but typically _at most_ one byte) is
reserved at the end of the memory space.

--
Peter
.

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