Two dimensional array Question

[mdh]

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]

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]

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".

 

[mdh]

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]

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[j] is an object of type T as long as 0<=i<=1 and
0<=j<=12. arr[j] is an element of arr but not element of arr.
arr is an element of arr and it is an array.


Remove del for email

 

[mdh]

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]

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]

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 and this is legal?
Thanks.

 

[Ben Bacarisse]

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 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;

means the same thing as

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

which can be simplified to

(*(arr + i) + 0)
*(arr + i)
arr

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

return 0;
}

 

[Ben Bacarisse]

pete <[email protected]> writes:

But it's better if you realize that

*p = arr;

means the same thing as

*p = &arr[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 != sizeof &arr[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;
means the same thing as
*p = (arr + 0);

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

Of course not, and that is one reason why I'd be loath to say that
arr and &arr[0] mean the same thing.

sizeof &arr[0]

*does* mean the same thing as

sizeof (&arr[0] + 0)

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

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

 

[Ben Bacarisse]

All true on the face of it but sizeof was not what was asked, it was
assignment.

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 Bacarisse]

Ben Bacarisse wrote:

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"

I didn't mean that.
I meant that given

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

then

*p = arr;

means exactly the same thing as

*p = &arr[0];

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

which can be simplified to

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

I hope I'd have kept out of it.

 

[Barry Schwarz]

 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 and this is legal?

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 will be converted to &arr[0] which is also of type T*.

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

 

[mdh]

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 will be converted to &arr[0] which is also of type T*.

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

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[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?

 

[Default User]

mdh wrote:

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

You do not reserve enough space for a null terminator. You will have
arrays of char but not strings in your array.

char *p;
char **q;

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

You should cast p to void*.

}

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);

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

 

[mdh]

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.

Point made and taken...thanks.

char arr [3][3] = {"One", "Two", "Lst"};

Legal, but dangerous. Don't treat these arrays of char as if they were
strings. They aren't.

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?

p=&arr[j];
printf( "%p\n", p);

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);

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.

You fell into the trap.
Yup!

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

Yes..I like it...thanks.

 

[mdh]

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.

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

 

[mdh]

Thanks Richard, as usual, for your insights.

 

[Chris Torek]

... strcpy can never initialise anything, because initialisation is what
you do to an object at the point where it is created.

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.

 

[Barry Schwarz]

int main (int argc, const char * argv[])
        {

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?