Velocity Reviews > Two dimensional array Question

# Two dimensional array Question

mdh
Guest
Posts: n/a

 05-08-2008
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.

Default User
Guest
Posts: n/a

 05-08-2008
Joe Wright wrote:

> 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.
>
> > 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!
> >

> No. arr[0] is (the address of) an array 13 of int.

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.

> 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.

Not to me.

p = &arr[0];

Would. Or what the OP had:

int *p;
p = arr[0];

Brian

Ben Bacarisse
Guest
Posts: n/a

 05-08-2008
mdh <(E-Mail Removed)> writes:

<When discussing int a[2][13];>

> 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.

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.

mdh
Guest
Posts: n/a

 05-09-2008
Thanks to all who responded. It makes a *little* more sense now, but I
think it will hopefully clear up with time and practice.

Barry Schwarz
Guest
Posts: n/a

 05-09-2008
On Thu, 8 May 2008 08:16:13 -0700 (PDT), mdh <(E-Mail Removed)> wrote:

>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.

The C99 standard uses the term "multidinensional array" so K&R is
slightly out of date in this regard.

>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.

What do you mean by "2". Is it an array of char arrays and the first
of such holds the string '2' '\0'?

>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

No. It is prohibited. arr[1] must immediately follow arr[0] in
memory.

>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.

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.

>
>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!

Yes and yes.

>
>
>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.

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

mdh
Guest
Posts: n/a

 05-09-2008
On May 8, 8:54*pm, Barry Schwarz <(E-Mail Removed)> wrote:
>
>
> The C99 standard uses the term "multidinensional array" so K&R is
> slightly out of date in this regard.
>
> >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.

>
> What do you mean by "2". *Is it an array of char arrays and the first
> of such holds the string '2' '\0'?

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....

>
> 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.
>

thank you Barry.

Default User
Guest
Posts: n/a

 05-09-2008
pete wrote:

> Default User wrote:
> > Joe Wright wrote:
> >
> > > 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.
> > >
> > > > 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!
> > > >
> > > No. arr[0] is (the address of) an array 13 of int.

> >
> > 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.

>
> You just made the same mistake that you just corrected.

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

mdh
Guest
Posts: n/a

 05-10-2008

>
>
> >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!

>
> Yes and yes.
>

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.

Ben Bacarisse
Guest
Posts: n/a

 05-10-2008
pete <(E-Mail Removed)> writes:

> mdh wrote:
>>>
>>>> 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!
>>> Yes and yes.
>>>

>> 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.

>
> 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];

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.

> because
>
> &arr[i][0] means &*(*(arr + i) + 0)
>
> which can be simplified to
>
> (*(arr + i) + 0)
> *(arr + i)
> arr[i]

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.

Ben Bacarisse
Guest
Posts: n/a

 05-10-2008
pete <(E-Mail Removed)> writes:

> Ben Bacarisse wrote:
>> pete <(E-Mail Removed)> writes:

<snip>
>>> But it's better if you realize that
>>>
>>> *p = arr[i];
>>>
>>> means the same thing as
>>>
>>> *p = &arr[i][0];

>>
>> 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).

>
> Think harder.

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".

> *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)

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.

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