incrementing a pointer to an array

Join Bytes!, India said:


I'm not sure. I don't know of any reason why it's illegal. Maybe Steve just
means it's a lousy idea.


Yes.

--
Richard Heathfield <http://www.cpax.org.uk>
Email: -http://www. +rjh@
Google users: <http://www.cpax.org.uk/prg/writings/googly.php>
"Usenet is a strange place" - dmr 29 July 1999

On Fri, 29 Aug 2008 23:29:22 -0700, Join Bytes!, India
wrote:

pointer to int */
The trick is here because you are getting the address of the pointer to
the first element, If I am not mistaken...

if you use sizeof on ap you will see that it is associated to the whole
array...8 bytes. Incrementing this address is undefined.

for intel compiler, the fist program prints
0 1
0
40896 <<<

utab said:


Wrong. ap is a pointer to the first element of the array, but using sizeof
on ap is by no means required to result in 8, and indeed on my system it
does not. Nor need it result in 3*sizeof(int) - and again, on my system it
does not.

#include <stdio.h>

int main(void)
{
int a1[3] = {0, 1, 2};
int (*ap)[3]; /* pointer to array [3] of int */
ap = &a1;
printf("**ap = %d\n", **ap);
printf("sizeof ap = %d\n", (int)sizeof ap);
ap++;
return 0;
}

**ap = 0
sizeof ap = 4



Why?

<snip>

--
Richard Heathfield <http://www.cpax.org.uk>
Email: -http://www. +rjh@
Google users: <http://www.cpax.org.uk/prg/writings/googly.php>
"Usenet is a strange place" - dmr 29 July 1999

On Sat, 30 Aug 2008 07:58:05 +0000, Richard Heathfield wrote:


Hi,

ap = &a1;

Could you clarify this statement for me? is not ap pointer to whole array
of 3?

not incrementing, dereferencing, sorry

utab said:


Yes, ap is a pointer to an array of three ints, and a1 is an array of three
ints, so it's perfectly legal to point ap at a1.


Right - but the OP has made it clear that he already knows dereferencing
such a pointer would be illegal. What he's unclear on is why the FAQ says
that incrementing it (just *one* place) is illegal.

Now that I've looked at the FAQ question in question, it seems to me that
this is simply a misunderstanding, and that the FAQ is at fault, but only
mildly so. What Steve is saying is that if your intent is to walk through
the array's objects, getting a pointer to the array isn't the way to do
it. Rather, you get a pointer to the first element in the array, and
increment /that/, in a loop - which is quite right, of course. His WRONG
applies to the error of treating a pointer-to-array as if it were a
pointer-to-first-element-of-array.

--
Richard Heathfield <http://www.cpax.org.uk>
Email: -http://www. +rjh@
Google users: <http://www.cpax.org.uk/prg/writings/googly.php>
"Usenet is a strange place" - dmr 29 July 1999

On Sat, 30 Aug 2008 08:19:07 +0000, Richard Heathfield wrote:


Right, my English ;) and fast read

In article <2e8c4eff-4156-4352-8279-a75f0e952657@z6g2000pre.googlegroups.com>,
Join Bytes!, India <subramanian100in@yahoo.comwrote:



Because the next line, which you have omitted, is

printf("%d\n", **ap); /* undefined */

The increment in isolation is not wrong, but if you're intending to
dereference the pointer, it is.

-- Richard
--
Please remember to mention me / in tapes you leave behind.

Join Bytes!, India wrote:

compile and run the following. Your implementation will certainly give
different values and perhaps a different format for the addresses. What
will none the less be true is that when ap = &a, ++ap is not the address
of any element of a. Why? Because ap is not a pointer to int, bur a
pointer to an array[3] of int, and the value of ++ap is the address of
the next array[3] of int.

#include <stdio.h>

int main(void)
{
int a[3] = { 0, 1, 2 };
int (*ap)[3]; /* pointer to array [3] of int */
size_t i, n = sizeof a / sizeof *a;
ap = &a;
printf("sizeof a = %zu, sizeof *a = %zu\n", sizeof a, sizeof *a);
for (i = 0; i < n; i++)
printf("&a[%zu] = %p\n", i, (void *) &a[i]);
putchar('\n');
printf("sizeof ap = %zu, sizeof *ap = %zu\n", sizeof ap,
sizeof *ap);
printf("ap = %p, ", (void *) ap);
printf("++ap = %p\n", (void *) ++ap);
return 0;
}


sizeof a = 12, sizeof *a = 4
&a[0] = dff9c
&a[1] = dffa0
&a[2] = dffa4

sizeof ap = 4, sizeof *ap = 12
ap = dff9c, ++ap = dffa8

Join Bytes!, India wrote:

It isn't WRONG taken by itself: It increments `ap'
to point one position past the thing it formerly pointed
at, which was the array `a1'. But it is WRONG when used
in combination with the next line in the FAQ's code, which
you snipped but which I now recommend to you for re-study.

--
Eric Sosman
Join Bytes!lid