About multidimensional array

[drazet]

hi,all

i have a question about multidimensional array.

int w[2][3],(*pw)[3];
pw=w;

is *(pw+1)[2] wrong?why?
and *(w[0]+2),pw[0][0] and *(pw[1]+2) mean?

Regards,
Drazet

[pete]

drazet wrote:

>
hi,all

i have a question about multidimensional array.

int w[2][3],(*pw)[3];
pw=w;

is *(pw+1)[2] wrong?why?
and *(w[0]+2),pw[0][0] and *(pw[1]+2) mean?

(A[X]) means the exact same thing as (*((A) + (X)))

--
pete

[CBFalconer]

drazet wrote:

>
i have a question about multidimensional array.

int w[2][3],(*pw)[3];
pw=w;

is *(pw+1)[2] wrong?why?
and *(w[0]+2),pw[0][0] and *(pw[1]+2) mean?

Consider the following display:

[1] c:\c\dispbase>explain int w[2][3]
declare w as array 2 of array 3 of int

[1] c:\c\dispbase>explain int (*pw)[3]
declare pw as pointer to array 3 of int

[1] c:\c\dispbase>alias explain
cdecl explain %1&

a pw probably can't hold any w. Use legitimate code first.

--
Chuck F (cbfalconer at maineline dot net)
Available for consulting/temporary embedded and systems.
<http://cbfalconer.home.att.net>
--
Posted via a free Usenet account from http://www.teranews.com

[Army1987]

On Mon, 20 Aug 2007 14:55:40 -0400, CBFalconer wrote:

drazet wrote:

>>
i have a question about multidimensional array.

int w[2][3],(*pw)[3];
pw=w;

is *(pw+1)[2] wrong?why?
and *(w[0]+2),pw[0][0] and *(pw[1]+2) mean?

Consider the following display:

[1] c:\c\dispbase>explain int w[2][3]
declare w as array 2 of array 3 of int

[1] c:\c\dispbase>explain int (*pw)[3]
declare pw as pointer to array 3 of int

[1] c:\c\dispbase>alias explain
cdecl explain %1&

a pw probably can't hold any w. Use legitimate code first.

w evaluates to &w[0], which is a pointer to array 3 of int. So
what's wrong with pw = w?
--
Army1987 (Replace "NOSPAM" with "email")
No-one ever won a game by resigning. -- S. Tartakower

[CBFalconer]

Army1987 wrote:

On Mon, 20 Aug 2007 14:55:40 -0400, CBFalconer wrote:

.... snip ...

>

>a pw probably can't hold any w. Use legitimate code first.

w evaluates to &w[0], which is a pointer to array 3 of int.
So what's wrong with pw = w?

It should be &w.

--
Chuck F (cbfalconer at maineline dot net)
Available for consulting/temporary embedded and systems.
<http://cbfalconer.home.att.net>

--
Posted via a free Usenet account from http://www.teranews.com

[pete]

CBFalconer wrote:

>
Army1987 wrote:

On Mon, 20 Aug 2007 14:55:40 -0400, CBFalconer wrote:

... snip ...

a pw probably can't hold any w. Use legitimate code first.

w evaluates to &w[0], which is a pointer to array 3 of int.
So what's wrong with pw = w?

It should be &w.

No.
w is implicitly converted to a pointer to
its first element, in this expression: (pw = w)
It's first element, is an array of 3 char.

Please try new.c with your compiler on strictest warning level.

/* BEGIN new.c */

int main(void)
{
int w[2][3], (*pw)[3];

pw = w;
return pw != w;
}

/* END new.c */

--
pete

[drazet]

Army1987 写�:

On Mon, 20 Aug 2007 14:55:40 -0400, CBFalconer wrote:

>drazet wrote:

>>i have a question about multidimensional array.

int w[2][3],(*pw)[3];
pw=w;

is *(pw+1)[2] wrong?why?
and *(w[0]+2),pw[0][0] and *(pw[1]+2) mean?

Consider the following display:

[1] c:\c\dispbase>explain int w[2][3]
declare w as array 2 of array 3 of int

[1] c:\c\dispbase>explain int (*pw)[3]
declare pw as pointer to array 3 of int

[1] c:\c\dispbase>alias explain
cdecl explain %1&

a pw probably can't hold any w. Use legitimate code first.

w evaluates to &w[0], which is a pointer to array 3 of int. So
what's wrong with pw = w?

pw = w is right,my question is that why the use of "*(w[0]+2),pw[0][0]
and *(pw[1]+2) "can get a right value,but "*(pw+1)[2]"gets a wrong?

[pete]

drazet wrote:

>
Army1987 写�:

On Mon, 20 Aug 2007 14:55:40 -0400, CBFalconer wrote:

drazet wrote:
i have a question about multidimensional array.

int w[2][3],(*pw)[3];
pw=w;

is *(pw+1)[2] wrong?why?
and *(w[0]+2),pw[0][0] and *(pw[1]+2) mean?
Consider the following display:

[1] c:\c\dispbase>explain int w[2][3]
declare w as array 2 of array 3 of int

[1] c:\c\dispbase>explain int (*pw)[3]
declare pw as pointer to array 3 of int

[1] c:\c\dispbase>alias explain
cdecl explain %1&

a pw probably can't hold any w. Use legitimate code first.

w evaluates to &w[0], which is a pointer to array 3 of int. So
what's wrong with pw = w?

pw = w is right,my question is that why the use of "*(w[0]+2),pw[0][0]
and *(pw[1]+2) "can get a right value,but "*(pw+1)[2]"gets a wrong?

It's supposed to be:

(*(pw+1))[2]
/* BEGIN new.c */

#include <stdio.h>

int main(void)
{
int w[2][3] = {1,2,3,4,5,6};
int (*pw)[3];

pw = w;
printf("%d\n", (*(pw+1))[2]);
return pw != w;
}

/* END new.c */

--
pete

[drazet]

pete 写�:

drazet wrote:

>Army1987 Ã¥â€*™éÂ�“:

>>On Mon, 20 Aug 2007 14:55:40 -0400, CBFalconer wrote:

drazet wrote:
i have a question about multidimensional array.
>
int w[2][3],(*pw)[3];
pw=w;
>
is *(pw+1)[2] wrong?why?
and *(w[0]+2),pw[0][0] and *(pw[1]+2) mean?
Consider the following display:

[1] c:\c\dispbase>explain int w[2][3]
declare w as array 2 of array 3 of int

[1] c:\c\dispbase>explain int (*pw)[3]
declare pw as pointer to array 3 of int

[1] c:\c\dispbase>alias explain
cdecl explain %1&

a pw probably can't hold any w. Use legitimate code first.
w evaluates to &w[0], which is a pointer to array 3 of int. So
what's wrong with pw = w?

pw = w is right,my question is that why the use of "*(w[0]+2),pw[0][0]
and *(pw[1]+2) "can get a right value,but "*(pw+1)[2]"gets a wrong?

It's supposed to be:

(*(pw+1))[2]
/* BEGIN new.c */

#include <stdio.h>

int main(void)
{
int w[2][3] = {1,2,3,4,5,6};
int (*pw)[3];

pw = w;
printf("%d\n", (*(pw+1))[2]);
return pw != w;
}

/* END new.c */

thanks a lot, i understand it,the (pw+1)[2] is the address of
w[3][0],but (*(pw+1))[2] is w[1][2].

[Alan Curry]

In article <fa**********@news.yaako.com>, drazet <ky******@gmail.comwrote:

>hi,all

i have a question about multidimensional array.

int w[2][3],(*pw)[3];
pw=w;

is *(pw+1)[2] wrong?why?

The [] has higher precedence than the * so this part:

(pw+1)[2]

is evaluated first. It's equivalent to these:

*((pw+1)+2)
*(pw+3)
pw[3]
w[3]

which is out of bounds.

(*(pw+1))[2] would be something different. Maybe it's what you meant.

--
Alan Curry
pa****@world.std.com

[somenath]

>
Then, given the definition that a[b] "means" (*(a) + (b)), this

Will it not be *((a) + (b)) ?

[CBFalconer]

drazet wrote:

>

.... snip about 140 lines ...

thanks a lot ~!~

Again, this answer, which was exactly as quoted, no less, obviously
required quoting all those 140 lines. Why can't people learn to
snip - its really quite easy.

--
Chuck F (cbfalconer at maineline dot net)
Available for consulting/temporary embedded and systems.
<http://cbfalconer.home.att.net>

--
Posted via a free Usenet account from http://www.teranews.com

[pete]

CBFalconer wrote:

>
drazet wrote:

... snip about 140 lines ...

thanks a lot ~!~

Again, this answer, which was exactly as quoted, no less, obviously
required quoting all those 140 lines. Why can't people learn to
snip - its really quite easy.

"Sometimes I do that as a cheap and easy way of saving
the explanation in my email "Sent box"."
--
pete

"Hey, I have a patent on that. Cease and desist immediately, or you
will hear from my lawyers."
--
Chuck F (cbfalconer at maineline dot net)

http://groups.google.com/group/comp....4?dmode=source

--
pete

[pete]

somenath wrote:

>

Then, given the definition that a[b] "means" (*(a) + (b)), this

Will it not be *((a) + (b)) ?

It is *((a) + (b)).

(a[b]) does not imply which of a or b is the pointer.

--
pete

[Richard Heathfield]

somenath said:

>

>>
Then, given the definition that a[b] "means" (*(a) + (b)), this

Will it not be *((a) + (b)) ?

Yes.

Congratulations! You caught out Chris Torek in a mistake. That is such a
rare achievement as to deserve a modding up.

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

[Richard Heathfield]

pete said:

somenath wrote:

>>

>
Then, given the definition that a[b] "means" (*(a) + (b)), this

Will it not be *((a) + (b)) ?

It is *((a) + (b)).

(a[b]) does not imply which of a or b is the pointer.

He didn't say it did. Read CT's upthread article again - it is indeed
incorrect, and that was a very good spot indeed from somenath.

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

[Richard]

pete <pf*****@mindspring.comwrites:

CBFalconer wrote:

>>
drazet wrote:

>

... snip about 140 lines ...

thanks a lot ~!~

Again, this answer, which was exactly as quoted, no less, obviously
required quoting all those 140 lines. Why can't people learn to
snip - its really quite easy.

"Sometimes I do that as a cheap and easy way of saving
the explanation in my email "Sent box"."
--
pete

"Hey, I have a patent on that. Cease and desist immediately, or you
will hear from my lawyers."
--
Chuck F (cbfalconer at maineline dot net)

http://groups.google.com/group/comp....4?dmode=source

So, the only contribution from Falconer is a wrong answer and a net
nanny. Good to see things don't change.

[Chris Torek]

[I wrote]

>Then, given the definition that a[b] "means" (*(a) + (b)), this

In article <11*********************@i38g2000prf.googlegroups. com>,
somenath <so*********@gmail.comwrote:

>Will it not be *((a) + (b)) ?

Oops, yes! Sorry, I managed to typo the placement of the "*" (or
miss out on a set of parentheses, although in this case I just
meant to put the "*" first).
--
In-Real-Life: Chris Torek, Wind River Systems
Salt Lake City, UT, USA (40°39.22'N, 111°50.29'W) [phone]
email: forget about it http://web.torek.net/torek/index.html
Reading email is like searching for food in the garbage, thanks to spammers.

[CBFalconer]

pete wrote:

CBFalconer wrote:

>drazet wrote:

... snip about 140 lines ...

>>thanks a lot ~!~

Again, this answer, which was exactly as quoted, no less, obviously
required quoting all those 140 lines. Why can't people learn to
snip - its really quite easy.

"Sometimes I do that as a cheap and easy way of saving
the explanation in my email "Sent box"."

Then don't. Most readers have a means of copying a response
somewhere. For example, I have an area named 'Infosave', with some
subdirectories. If I need to keep a message I just copy it there.
Takes about 2 1/4 seconds.

You are using Mozilla, which is a descendent of the Netscape I am
using. Investigate the message menu, especially the copy submenu.

drazet is using Thunderbird, and the same suggestion applies.

--
Chuck F (cbfalconer at maineline dot net)
Available for consulting/temporary embedded and systems.
<http://cbfalconer.home.att.net>

--
Posted via a free Usenet account from http://www.teranews.com

[CBFalconer]

Richard Heathfield wrote:

somenath said:

>>Then, given the definition that a[b] "means" (*(a) + (b)), this

Will it not be *((a) + (b)) ?

Yes.

Congratulations! You caught out Chris Torek in a mistake. That is
such a rare achievement as to deserve a modding up.

Does Chris Torek hand out checks (or cheques) for $2.56?

--
Chuck F (cbfalconer at maineline dot net)
Available for consulting/temporary embedded and systems.
<http://cbfalconer.home.att.net>

--
Posted via a free Usenet account from http://www.teranews.com

[Richard Heathfield]

CBFalconer said:

Richard Heathfield wrote:

>somenath said:

>>>Then, given the definition that a[b] "means" (*(a) + (b)), this

Will it not be *((a) + (b)) ?

Yes.

Congratulations! You caught out Chris Torek in a mistake. That is
such a rare achievement as to deserve a modding up.

Does Chris Torek hand out checks (or cheques) for $2.56?

Not as far as I'm aware.

I would not have spotted somenath's correction, were it not for pete's
reply to him, which - unusually for pete - missed the point completely.
That's because somenath uses a gmail account, and by default I ignore
articles coming from such accounts. I have modded up several gmail
users known to be clueful, however, and he has now joined their ranks.

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

[pete]

CBFalconer wrote:

>
pete wrote:

CBFalconer wrote:

You are using Mozilla,
which is a descendent of the Netscape I am using.

Is Mozilla 3.04Gold really descended from Mozilla 4.75 ?

From: CBFalconer <cb********@yahoo.com>
X-Mailer: Mozilla 4.75 [en] (Win98; U)

From: pete <pf*****@mindspring.com>
X-Mailer: Mozilla 3.04Gold (WinNT; I)

Investigate the message menu, especially the copy submenu.

I have no copy submenu.

--
pete

[CBFalconer]

pete wrote:

CBFalconer wrote:

>You are using Mozilla,
which is a descendent of the Netscape I am using.

Is Mozilla 3.04Gold really descended from Mozilla 4.75 ?

Woops - somebody is using older software than I am!

--
Chuck F (cbfalconer at maineline dot net)
Available for consulting/temporary embedded and systems.
<http://cbfalconer.home.att.net>

--
Posted via a free Usenet account from http://www.teranews.com

[Barry Schwarz]

On Mon, 20 Aug 2007 17:43:13 +0800, drazet <ky******@gmail.comwrote:

>hi,all

i have a question about multidimensional array.

int w[2][3],(*pw)[3];
pw=w;

With a few exceptions that don't apply to your examples, an expression
of type array is converted to the address of the first element of the
array with type pointer to element type. So this statement is exactly
the same as pw = &w[0].

>
is *(pw+1)[2] wrong?why?

Since [] has higher precedence than *, this is parsed as
*((pw+1)[2])

(pw+1) evaluates to the address of w[1].

(pw+1)[0] evaluates to the array (pw+1) points to, namely w[1].
(pw+1)[1] would evaluate to the next array beyond w[1] and (pw+1)[2]
would evaluate to the second array beyond w[1]. Unfortunately,
neither of these two arrays exist since w[0] and w[1] are the only two
defined.

If w[3] existed, then *(pw+1)[2] would evaluate to w[3][0].

If your intent was to get the last scalar element of w, then use
either (*(pw+1))[2] or the equivalent and preferable pw[1][2].

>and *(w[0]+2),pw[0][0] and *(pw[1]+2) mean?

By definition, *(w[0]+2) is exactly the same as w[0][2].

Since pw points to w, pw[0][0] is the same as w[0][0].

Applying the previous two rules, you should have no trouble
deciphering *(pw[1]+2).
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