About multidimensional array

[CBFalconer]

"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.

 

[CBFalconer]

somenath said:

Then, given the definition that a "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?

 

[Richard Heathfield]

CBFalconer said:

somenath said:

Then, given the definition that a "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.

 

[pete]

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 <[email protected]>
X-Mailer: Mozilla 4.75 [en] (Win98; U)

Investigate the message menu, especially the copy submenu.

I have no copy submenu.

 

[CBFalconer]

Is Mozilla 3.04Gold really descended from Mozilla 4.75 ?

Woops - somebody is using older software than I am!

 

[Barry Schwarz]

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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