Two dimensional array Question

I am still having a problem understanding K&RII on p 112. I have
looked at the FAQs --which I am sure answer it in a way that I have
missed, so here goes.
A 2-dim array, (per K&R) is really a 1-dim array, each of whose
elements is an array.
Looking at the debugger I use, arr[2][13] is shown as an initial value
of "2", but when expanded, there are 2 consecutive arrays of 13
elements.
So, may I ask this?
Is there anything special that marks the end of the first 13 elements
from the beginning of the 2nd 13 elements? In other words, it seems to
me that this is nothing more than a 1-dim array, with the information
about the structure of the array provided by the declaration?
Hopefully this makes sense.

The exercise associated with this, used the construct

*p = arr[1 or 0 ] to point to either the "first" or "second" row of
the array. Does the compiler "know" where to point to because it has
been given this information by the declaration of "13" in arr[2][13].
Hopefully this makes sense too!


And lastly, K&R's description of "each of whose elements is an array"
has never made sense to me. It may be that the answer to the above may
clarify it.

Thanks in advance.

re: Two dimensional array Question

mdh wrote:Perfect sense. All above is correct. Given 'int arr[2][13];', arr is an
array 2 of array 13 of int. 26 consecutive ints.
No. arr[0] is (the address of) an array 13 of int. The pointer 'p' must
be declared to reflect this.

int (*p)[13];

Now p is a pointer to array 13 of int and..

p = arr[0];

...makes sense.I hope so. You're really not too far off. Hang in there.
--
Joe Wright
"Everything should be made as simple as possible, but not simpler."
--- Albert Einstein ---

re: Two dimensional array Question

Joe Wright wrote:
No, it's not. It is an array 13 of int, not a pointer to one. arr is a
pointer to array 13 of int.
Not to me.

p = &arr[0];

Would. Or what the OP had:

int *p;
p = arr[0];





Brian

re: Two dimensional array Question

mdh <mdeh@comcast.netwrites:

<When discussing int a[2][13];>
Another thought... Do you have a problem with:

struct ints { int a,b,c,d,e,f,g,h,i,j,k,l,m; };
struct ints a[2];

being an array, each of whose elements is a structure of 13 ints?
What if we say:

struct ints { int e[13]; };
struct ints a[2];

For many people the trouble is the way it is written. If it were
something like: 'int[13] a[2];' they might find it easier to swallow
but the array declarators group like this:

int (a[2]) [13] ;

(you can write that if you want!) so a is "an array of 2... arrays of
13... ints. I think you are very close to "getting it".

--
Ben.

re: Two dimensional array Question

Default User wrote:You just made the same mistake that you just corrected.

arr isn't any more of a pointer than arr[0] is.
arr isn't any less of an array than arr[0] is.

(sizeof arr) equals (2 * 13 * sizeof(int)).

--
pete

re: Two dimensional array Question

Thanks to all who responded. It makes a *little* more sense now, but I
think it will hopefully clear up with time and practice.

re: Two dimensional array Question

On Thu, 8 May 2008 08:16:13 -0700 (PDT), mdh <mdeh@comcast.netwrote:
The C99 standard uses the term "multidinensional array" so K&R is
slightly out of date in this regard.
What do you mean by "2". Is it an array of char arrays and the first
of such holds the string '2' '\0'?
No. It is prohibited. arr[1] must immediately follow arr[0] in
memory.
It is true that the 26 "basic elements" of arr will appear in memory
"linearly". This has prompted many debates about whether a pointer to
arr[0][0] can be used with subscripts ranging from 0 to 25 to access
all 26. What is not in dispute is that in a 1d array T x[N], x[0]
evaluates to an element of type T. In your case, arr[0] evaluates to
an array of T. To my mind, this is sufficient to say arr cannot be
considered a 1d array.
Yes and yes.
arr[0] is an array 13 T. arr[1] is an array of 13 T. arr[2] does not
exist. arr[i][j] is an object of type T as long as 0<=i<=1 and
0<=j<=12. arr[i][j] is an element of arr[i] but not element of arr.
arr[i] is an element of arr and it is an array.


Remove del for email

re: Two dimensional array Question

On May 8, 8:54*pm, Barry Schwarz <schwa...@dqel.comwrote:
I am using Xcode...not that that probably makes any difference.
Initially arr[2][13] is declared as a char array. In the debugger
window, there is a column for value and the "initial" value of arr is
"2". If one expands this, then one sees 2 arrays of 13 elements, and
as you say below, these seem to be contiguous. I was trying to find a
way of seeing what the address of each element is, but have been
unable to do so....


thank you Barry.

re: Two dimensional array Question

pete wrote:
Yes, I did in a way. I was sloppy about context. It is correct that arr
is not a pointer, although it's converted to one in most contexts.

The code was correct, which it wasn't originally.


Thanks for clarifying my stab at i.





Brian

re: Two dimensional array Question

I was in a hurry to get to work this am...so forgot to ask this.
In a multidimensional array, one can simply drop the second []? as in
*p=arr[i] and this is legal?
Thanks.

re: Two dimensional array Question

mdh wrote:It's legal regardless of whether or not
you understand what you're writing.
But it's better if you realize that

*p = arr[i];

means the same thing as

*p = &arr[i][0];

because

&arr[i][0] means &*(*(arr + i) + 0)

which can be simplified to

(*(arr + i) + 0)
*(arr + i)
arr[i]

--
pete

re: Two dimensional array Question

pete <pfiland@mindspring.comwrites:
The trouble is this phrase "means the same as". Both right hand sides
are certainly closely related expressions, but they have different
types, different sizes (sizeof arr[i] != sizeof &arr[i][0]) and they
can't be used in the same places (you can, for example take the
address of one but not of the other).

Sometimes it helps to gloss over these differences and sometimes it
does not. I am not sure which is the case here.
C is not good for equational reasoning. The OP might find this
helpful:

#include <stdio.h>

#define PSZ(exp) printf("sizeof " #exp " = %d\n", (int)sizeof exp)

int main(void)
{
int i = 0;
int arr[2][13] = {0};

PSZ(&arr[i][0]);
PSZ(&*(*(arr + i) + 0));
PSZ((*(arr + i) + 0));
PSZ(*(arr + i));
PSZ(arr[i]);

return 0;
}

--
Ben.

re: Two dimensional array Question

Ben Bacarisse wrote:Think harder.

*p = arr[i];
means the same thing as
*p = (arr[i] + 0);

sizeof arr[i]
does not mean the same thing as
sizeof (arr[i] + 0)

My claim was about two expression statements.
You're talking about array type expressions
in a context which is irrelevant to this.

Consider new.c:
Do you think new.c is missing a cast somewhere?
Do you have a compiler that thinks
that new.c is missing a cast somewhere?
Do you know under what circumstances
cast operators are required in C?

/* BEGIN new.c */

int main(void)
{
int arr[2][13] = {0};
int **p1;
int **p2;

*p1 = arr[1];
*p2 = &arr[1][0];
return *p1 != *p2;
}

/* END new.c */


--
pete

re: Two dimensional array Question

Ben Bacarisse wrote:All true on the face of it but sizeof was not what was asked, it was
assignment. Given..

int *p;
p = arr[i];

Yes, arr[i] is an array 13 of int but expressing an array as above
yields the address of its first element, in this case int*.

p = &arr[i][0];

The address of the int yields the same int*.

--
Joe Wright
"Everything should be made as simple as possible, but not simpler."
--- Albert Einstein ---

re: Two dimensional array Question

pete <pfiland@mindspring.comwrites:
<snip>OK, but you'll have to help me if you think I've missed something. I
really do think I get it.

I know you meant "in most expression contexts" and I did not say you
said anything wrong. All I was saying is that it sometimes helps to
know the ways in which they *don't* mean the same thing. Obviously
you disagree, and that is fine. Only the OP can say which is more
helpful in "getting it".
Of course not, and that is one reason why I'd be loath to say that
arr[i] and &arr[i][0] mean the same thing.

sizeof &arr[i][0]

*does* mean the same thing as

sizeof (&arr[i][0] + 0)

(as do '*p = &arr[i][0];' and '*p = &arr[i][0] + 0;').

All this does is to highlight that we expect different things from two
expressions that "mean the same".

--
Ben.

re: Two dimensional array Question

Joe Wright <joewwright@comcast.netwrites:
<snip>Yes, I know I took it out of context. It was deliberate because I
wanted to point out that context was critical. Maybe I complicated
matters and, if so, I am sorry.

--
Ben.

re: Two dimensional array Question

Ben Bacarisse wrote:I didn't mean that.
I meant that given

int arr[2][13], **p, i;

then

*p = arr[i];

means exactly the same thing as

*p = &arr[i][0];


--
pete

re: Two dimensional array Question

pete <pfiland@mindspring.comwrites:
<snip>Gotcha. You were making a very narrow point.

Sadly, you led me a little up the path I ended up running down.
Rather than stopping there, you went on to expand &arr[i][0] in stages
(without the *p = context) which I jumped on as implying a more
general equivalence.

If you'd kept the context and written:

"*p = &arr[i][0] means *p = &*(*(arr + i) + 0)

which can be simplified to

*p = (*(arr + i) + 0)
*p = *(arr + i)
*p = arr[i]

I hope I'd have kept out of it.

--
Ben.

re: Two dimensional array Question

On May 9, 8:35*pm, mdh <m...@comcast.netwrote:
Depending on the type of p. If arr is defined as T arr[N][M], then
for your code to be legal, p must be a T**. *p will then have type T*
and arr[i] will be converted to &arr[i][0] which is also of type T*.

Far more common is for p to be a T* and the assignment to read
p = arr[i].

re: Two dimensional array Question

On May 11, 4:17*am, Barry Schwarz <schwar...@yahoo.comwrote:
Well...I **think** I see it a little more clearly.
Although once admonished for writing code to figure things out, a weak
attempt follows.

#include <stdio.h>

int main (int argc, const char * argv[])
{
int i, j, k, l;
char arr [3][3] = {"One", "Two", "Lst"};
char *p;
char **q;

for ( i = 0; i < 3; i ++)
for ( j= 0; j < 3; j++)
{
p=&arr[i][j];
printf( "%p\n", p);
}

putchar('\n');

for ( k = 0; k < 3; k ++)
for ( l= 0; l < 3; l++)
{
p=arr[k];
q=&arr[k][l];
printf( "%s %s\n", p, q);
}
}
output:

0xbffff7cf
0xbffff7d0
0xbffff7d1
0xbffff7d2
0xbffff7d3
0xbffff7d4
0xbffff7d5
0xbffff7d6
0xbffff7d7

OneTwoLst OneTwoLst
OneTwoLst neTwoLst
OneTwoLst eTwoLst
TwoLst TwoLst
TwoLst woLst
TwoLst oLst
Lst Lst
Lst st
Lst t

Which I think says, as mentioned above, that the addresses are
contiguous. So at least that is an easy visual and conceptual concept
to comprehend.
And to repeat what you put far more elegantly, more to see if I am
understanding it,
char *p in the context of the declaration arr[m][n], is a "row"
and char **q, is a character?

re: Two dimensional array Question

mdh said:

<snip>
Writing code "to figure things out" is a two-edged sword. It can certainly
be helpful for discovering how things *appear* to work, but it is a poor
guide for discovering how things *must* work. For example, the way a
for-loop works is well-defined, so if we've forgotten whether the
increment step (it isn't necessarily an increment, but let's keep things
simple) of a for-loop is always executed at least once, we can write a
quick program to find out:

#include <stdio.h>

int main(void)
{
int i;
int j = 0;
for(i = 0; i < j; i++)
{
printf("i is now %d\n");
}
printf("Outside the loop, i is now %d\n");
return 0;
}

and this is perfectly okay. What we /can't/ do is take a program like this:

#include <stdio.h>

int main(void)
{
printf("%d\n", 'A');
return 0;
}

and deduce from its output that the value of 'A' is *required* to be 193.
After all, you might not be running this program on an EBCDIC system.

So we have to be careful, that's all. Now, to your question:
Legal, but dangerous. Don't treat these arrays of char as if they were
strings. They aren't.
Although the representations of char * and void * are required to be the
same, it is generally wiser to insist on void * for %p:

printf("%p\n", (void *)p);
You fell into the trap. Neither p nor q points at a string.
....is irrelevant, since the program invokes undefined behaviour by treating
non-strings as if they were strings.

<snip>
The easiest way to comprehend multidimensional arrays is to think of them
as arrays of arrays. If you never try to breach the bounds of an object,
all will be well. By treating arrays-of-arrays as if they were a single
contiguous object (EVEN IF THAT IS TRUE), you're violating that principle.
Now, there are times when it can be necessary to do that (for example,
inspecting object representations), but one should tread very carefully
when so doing.
Not quite. In your first loop you point p to an individual character, so
that's what it points at. In your second loop, you point p to the first
element in an array, so it's reasonable to think of p as pointing at a
row.

q is only used in the second loop, where it most certainly is not a
character but yes, it does point at one.

--
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

re: Two dimensional array Question

[sigh]

Richard Heathfield said:
Spot both...
....mistakes. :-(
....after fixing!

--
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

re: Two dimensional array Question

mdh wrote:

You do not reserve enough space for a null terminator. You will have
arrays of char but not strings in your array.
You should cast p to void*.
You attempt to pass the non-strings when printf() is expecting strings.
Your program now exhibits undefined behavior. Nothing else about it
will tell you very much.



Brian

re: Two dimensional array Question

Richard Heathfield wrote:printf("i is now 0\n");
printf("Outside the loop, i is now 0\n");

--
pete

re: Two dimensional array Question

On May 11, 8:16 pm, Richard Heathfield <r...@see.sig.invalidwrote:
Point made and taken...thanks.

OK..now let me show my total ignorance.

So, can multidim char arrays ever be treated as strings and would
this then be the correct intialization? char arr[] [] = "one, two";
And, for a regular (one dim) char array, would this be a string
intialization? char arr[]= "something" or the use of strcpy?

And, somewhat philosophically, a (one dim) char array is simply a
contiguous set of bytes. If one ? *knows* the array is NULL
terminated, then we can safely treat it as a string. But the *only*
way to know this is if it is correctly initialized in the first place.
Is this a fair way of looking at it?


I did read today about the need for a pointer to be cast to
void ...but is there a reason for this? Other than the language says
it should be so.

Yup!
Yes..I like it...thanks.

re: Two dimensional array Question

Thanks Brian...I think I am beginning to understand this.Not quite
there yet...but getting closer!! Next onto function pointers!!!

re: Two dimensional array Question

mdh said:
Yes. There is *nothing special* about "multidim" arrays. For example:

char arr[6][15] = { "This", "string", "initialisation",
"works", "just", "fine" };
for(i = 0; i < 5; i++)
{
printf("%s\n", arr[i]);
}
All arrays are regular. A multidimensional array is simply a
one-dimensional array, each of whose elements happens to be an array.
Divide and conquer.
Yes.
No, strcpy can never initialise anything, because initialisation is what
you do to an object at the point where it is created. It says to the
compiler: "at the point where you are making an object of THIS type for
me, please give it THIS value". By the time you can call strcpy, it's too
late to initialise the array into which you are copying the string, but of
course it's not too late to update that array's members' values.
A char array cannot be NULL terminated. C is a case sensitive language.
NULL is a null pointer constant. Strings are terminated by a null
character, not a NULL character (because there is no such thing as a NULL
character).
Right.
Yes. When you create an object, put something in it, so that it has a known
value. That way, later on when you examine that object, you know you're
/allowed/ to examine that object. It might not have the /right/ thing in
it (i.e. your program might have a bug) but at least it contains a
legitimate and reproducible value (i.e. debugging is made considerably
easier).
Yes. The printf function specifies an interface that allows you to print a
void * value, but doesn't specify an interface for pointer values of other
types. If you would like a function that can print a pointer value of
*any* type, feel free to write one - but I should warn you that it will
take you infinitely long to write such a function in true generality.
Isn't that a pretty powerful reason?

--
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

re: Two dimensional array Question

Thanks Richard, as usual, for your insights.

re: Two dimensional array Question

In article <Z8adnbR2Ut4wgbTVnZ2dnUVZ8rednZ2d@bt.com>
Richard Heathfield <rjh@see.sig.invalidwrote:In the sense of "an initializer" (as used in Standard C), yes.
However, one should be aware that many (perhaps even most?) people
also refer to the first assignment to something as its
"initialization", and then use the word "initialize" (or
"initialise") to describe what happens at that point.

That is:

void f(void) {
int i = 0;
int j;

... code that does not touch j ...
j = 0;
...
}

In Standard C terminology, the variable i has an initializer and
is initialized to 0. The variable j has no initializer and is not
initialized. The assignment "j = 0" is an ordinary assignment and
does not "initialize" j ... but in many/most people's terminology,
the assignment to j, which is its initial (i.e., first) assignment,
does "initialize" j.

In short, one must be careful with terminology, as the multiple
meanings of various words do confuse people. :-)
--
In-Real-Life: Chris Torek, Wind River Systems
Salt Lake City, UT, USA (40°39.22'N, 111°50.29'W) +1 801 277 2603
email: gmail (figure it out) http://web.torek.net/torek/index.html

re: Two dimensional array Question

Chris Torek said:

<snip>
Right.
Well, Chris, I can't help that. :-)
That's why I try very hard to stick to the terminology defined by ISO.
Rather than people having to guess what I mean when I use a term, they
have the opportunity to look it up in the Standard.

Of course, not all terms are defined therein, and even when they are I
don't suppose I always get them right. One does the best one can.

--
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

re: Two dimensional array Question

On May 11, 7:07*pm, mdh <m...@comcast.netwrote:Did you throw the const in just to annoy people? Is that "signature"
properly documented on your system? How many other systems do you
think will support it?

re: Two dimensional array Question

mdh wrote:A char arr[3] = "One";
B char arr[] = {'O','n','e'};

C char arr[4] = "One";
D char arr[] = {'O','n','e','\0'};
E char arr[] = "One";

Array definitions A and B, can be used interchangeably.
Array definitions C, D and E, can be used interchangeably.

--
pete