Review: 3D-Array

[Vijay Kumar R Zanvar]

Hi c.l.c,

Is this a correct method of allocating a 3-dimensional
array?

/* assume all malloc's succeed */
#define MAT 2
#define ROW 2
#define COL 2

char ***ptr = malloc ( MAT * sizeof *ptr );

for ( i = 0; i < MAT; i++ )
ptr[i] = malloc ( ROW * sizeof **ptr );

for ( i = 0; i < MAT; i++ )
for ( j = 0; j < ROW; j++ )
ptr[i][j] = malloc ( COL * sizeof ***ptr );

Thanks.

--
"When you will be pleased to dine, Mr. Homles?" asked Mrs Hudson,
the landlady, curiosly.
"At seven thirty, the day after tomorrow." said he invovled in his work.

[Sean Kenwrick]

"Vijay Kumar R Zanvar" <vi*****@hotpop.com> wrote in message
news:bv************@ID-203837.news.uni-berlin.de...

Hi c.l.c,

Is this a correct method of allocating a 3-dimensional
array?

/* assume all malloc's succeed */
#define MAT 2
#define ROW 2
#define COL 2

char ***ptr = malloc ( MAT * sizeof *ptr );

for ( i = 0; i < MAT; i++ )
ptr[i] = malloc ( ROW * sizeof **ptr );

for ( i = 0; i < MAT; i++ )
for ( j = 0; j < ROW; j++ )
ptr[i][j] = malloc ( COL * sizeof ***ptr );

Thanks.

You define a pointer to a pointer to a pointer to a char. So your first
malloc() should create the space for several 'pointer to a pointer to a
char' and make ptr (which is a pointer to a "pointer to a pointer to char"
point to it through the malloc call)

char *** ptr=malloc(MAT * sizeof **ptr);

For each of the entries in this array of pointers to pointers to a char you
now need to point to a 'pointer to a char' type:

for(i=0;i<MAT;i++)
ptr[i]=malloc(ROW * sizeof *ptr);

So now you have a bunch of ROW pointers to chars in each row of the array
ptr[i] so now you must point these to the 'char' (or the first char in an
array of char)

for ( i = 0; i < MAT; i++ )
for ( j = 0; j < ROW; j++ )
ptr[i][j] = malloc ( COL * sizeof char );
You can think of it more clearly like this:

char array[MAT][ROW][COL];

char * array[MAT][ROW]

char ** array[MAT]

char *** array;

In practice the sizeof ***ptr and sizeof ** ptr etc will all be the same.
Therefore what you end up with in your original code is a three dimentional
array of the type char * as follows:

char * array[MAT][ROW][COL];

Which might be what you wanted, but all the char * array entries you created
do not yet point anywhere and when you do point them somewhere you will in
fact end up with a 4 dimensional array like this for example:

char array[MAT][ROW][COL][STRLEN+1];

Hope this helps you clarify this...

Sean

[Thes]

Sean Kenwrick wrote:

"Vijay Kumar R Zanvar" <vi*****@hotpop.com> wrote in message
news:bv************@ID-203837.news.uni-berlin.de...

Hi c.l.c,

Is this a correct method of allocating a 3-dimensional
array?

/* assume all malloc's succeed */
#define MAT 2
#define ROW 2
#define COL 2

char ***ptr = malloc ( MAT * sizeof *ptr );

for ( i = 0; i < MAT; i++ )
ptr[i] = malloc ( ROW * sizeof **ptr );

for ( i = 0; i < MAT; i++ )
for ( j = 0; j < ROW; j++ )
ptr[i][j] = malloc ( COL * sizeof ***ptr );

Thanks.

You define a pointer to a pointer to a pointer to a char. So your first
malloc() should create the space for several 'pointer to a pointer to a
char' and make ptr (which is a pointer to a "pointer to a pointer to char"
point to it through the malloc call)

char *** ptr=malloc(MAT * sizeof **ptr);

Are you sure about that? Isn't sizeof **ptr equivalent to sizeof char*
seeing as ptr is of type char***? Perhaps you are confusing "**ptr" with
"char**"?

I think you should use
char*** ptr = malloc(MAT * sizeof *ptr);
here instead.


For each of the entries in this array of pointers to pointers to a char you
now need to point to a 'pointer to a char' type:

for(i=0;i<MAT;i++)
ptr[i]=malloc(ROW * sizeof *ptr);
Similarly "*ptr" is not the same as "char*". Use:
ptr[i] = malloc(ROW * sizeof *(ptr[i]));

So now you have a bunch of ROW pointers to chars in each row of the array
ptr[i] so now you must point these to the 'char' (or the first char in an
array of char)

for ( i = 0; i < MAT; i++ )
for ( j = 0; j < ROW; j++ )
ptr[i][j] = malloc ( COL * sizeof char );
And here:
ptr[i][j] = malloc(COL * sizeof *(ptr[i][j]));


You can think of it more clearly like this:

char array[MAT][ROW][COL];

char * array[MAT][ROW]

char ** array[MAT]

char *** array;

In practice the sizeof ***ptr and sizeof ** ptr etc will all be the same.
Therefore what you end up with in your original code is a three dimentional
array of the type char * as follows:

char * array[MAT][ROW][COL];
I'd have though it was more like char array[MAT][ROW][COL];
The array has three sets of braces so it decays to a triple pointer
right? Which is exactly what the OP started with.


Which might be what you wanted, but all the char * array entries you created
do not yet point anywhere and when you do point them somewhere you will in
fact end up with a 4 dimensional array like this for example:

char array[MAT][ROW][COL][STRLEN+1];

Hope this helps you clarify this...

Sean

Hmmm - I'm a bit confused by all this...

[nrk]

Vijay Kumar R Zanvar wrote:

Hi c.l.c,

Is this a correct method of allocating a 3-dimensional
array?

/* assume all malloc's succeed */
#define MAT 2
#define ROW 2
#define COL 2

char ***ptr = malloc ( MAT * sizeof *ptr );

for ( i = 0; i < MAT; i++ )
ptr[i] = malloc ( ROW * sizeof **ptr );

for ( i = 0; i < MAT; i++ )
for ( j = 0; j < ROW; j++ )
ptr[i][j] = malloc ( COL * sizeof ***ptr );

Thanks.

Correct. Note that as someone who's posted here more than once before, you
must know this is answered in the FAQ. Specifically Q6.16:
http://www.eskimo.com/~scs/C-faq/q6.16.html

-nrk.

--
Remove devnull for email

[nrk]

Sean Kenwrick wrote:

"Vijay Kumar R Zanvar" <vi*****@hotpop.com> wrote in message
news:bv************@ID-203837.news.uni-berlin.de...

Hi c.l.c,

Is this a correct method of allocating a 3-dimensional
array?

/* assume all malloc's succeed */
#define MAT 2
#define ROW 2
#define COL 2

char ***ptr = malloc ( MAT * sizeof *ptr );

for ( i = 0; i < MAT; i++ )
ptr[i] = malloc ( ROW * sizeof **ptr );

for ( i = 0; i < MAT; i++ )
for ( j = 0; j < ROW; j++ )
ptr[i][j] = malloc ( COL * sizeof ***ptr );

Thanks.
You define a pointer to a pointer to a pointer to a char. So your
first malloc() should create the space for several 'pointer to a pointer
to a
char' and make ptr (which is a pointer to a "pointer to a pointer to
char" point to it through the malloc call)

char *** ptr=malloc(MAT * sizeof **ptr);

Wrong. OP was correct. **ptr has type pointer-to-char, and you need sizeof
pointer-to-pointer-to-char.
For each of the entries in this array of pointers to pointers to a char
you now need to point to a 'pointer to a char' type:

for(i=0;i<MAT;i++)
ptr[i]=malloc(ROW * sizeof *ptr);

Wrong again. *ptr has type pointer-to-pointer-to-char, and you need sizeof
pointer-to-char.
So now you have a bunch of ROW pointers to chars in each row of the array
ptr[i] so now you must point these to the 'char' (or the first char in an
array of char)

for ( i = 0; i < MAT; i++ )
for ( j = 0; j < ROW; j++ )
ptr[i][j] = malloc ( COL * sizeof char );

Ok. But I would prefer:
ptr[i][j] = malloc(COL * sizeof ***ptr);

You can think of it more clearly like this:

char array[MAT][ROW][COL];

char * array[MAT][ROW]

char ** array[MAT]

char *** array;

In practice the sizeof ***ptr and sizeof ** ptr etc will all be the same.
Absolute nonsense, particularly given OP's declaration of ptr. Don't bandy
about such dangerous and incorrect advice.
Therefore what you end up with in your original code is a three
dimentional array of the type char * as follows:

char * array[MAT][ROW][COL];

OP wants a two dimensional array of pointer-to-char.
Which might be what you wanted, but all the char * array entries you
created
do not yet point anywhere and when you do point them somewhere you will
in fact end up with a 4 dimensional array like this for example:

char array[MAT][ROW][COL][STRLEN+1];

Wrong again. OP's code is fine.
Hope this helps you clarify this...

No. It certainly doesn't.

-nrk.
Sean

--
Remove devnull for email

[Sean Kenwrick]

"nrk" <ra*********@devnull.verizon.net> wrote in message
news:kd**************@nwrddc02.gnilink.net...

Sean Kenwrick wrote:

"Vijay Kumar R Zanvar" <vi*****@hotpop.com> wrote in message
news:bv************@ID-203837.news.uni-berlin.de...

Hi c.l.c,

Is this a correct method of allocating a 3-dimensional
array?

/* assume all malloc's succeed */
#define MAT 2
#define ROW 2
#define COL 2

char ***ptr = malloc ( MAT * sizeof *ptr );

for ( i = 0; i < MAT; i++ )
ptr[i] = malloc ( ROW * sizeof **ptr );

for ( i = 0; i < MAT; i++ )
for ( j = 0; j < ROW; j++ )
ptr[i][j] = malloc ( COL * sizeof ***ptr );

Thanks.
You define a pointer to a pointer to a pointer to a char. So your
first malloc() should create the space for several 'pointer to a pointer
to a
char' and make ptr (which is a pointer to a "pointer to a pointer to
char" point to it through the malloc call)

char *** ptr=malloc(MAT * sizeof **ptr);

Wrong. OP was correct. **ptr has type pointer-to-char, and you need

sizeof pointer-to-pointer-to-char.
Yeh, I meant to use the TYPE not the object here so it would be:

char *** ptr=malloc(MAT * sizeof (char**));

Hence the two '**' which was wrong when I used them with sizeof **ptr...

For each of the entries in this array of pointers to pointers to a char
you now need to point to a 'pointer to a char' type:

for(i=0;i<MAT;i++)
ptr[i]=malloc(ROW * sizeof *ptr);

Wrong again. *ptr has type pointer-to-pointer-to-char, and you need

sizeof pointer-to-char.
Again I meant 'char *' not *ptr - very careless...

So now you have a bunch of ROW pointers to chars in each row of the array ptr[i] so now you must point these to the 'char' (or the first char in an array of char)

for ( i = 0; i < MAT; i++ )
for ( j = 0; j < ROW; j++ )
ptr[i][j] = malloc ( COL * sizeof char );

Ok. But I would prefer:
ptr[i][j] = malloc(COL * sizeof ***ptr);

You can think of it more clearly like this:

char array[MAT][ROW][COL];

char * array[MAT][ROW]

char ** array[MAT]

char *** array;

In practice the sizeof ***ptr and sizeof ** ptr etc will all be the

same.
Absolute nonsense, particularly given OP's declaration of ptr. Don't bandy about such dangerous and incorrect advice.

Again I meant sizeof(char ***) and sizeof(char **) would be the same in
practice ...

I think my brain suffered a temporary blue screen of death...

Sean

[Arthur J. O'Dwyer]

On Fri, 30 Jan 2004, Sean Kenwrick wrote:

"nrk" <ra*********@devnull.verizon.net> wrote...

Sean Kenwrick wrote:

You define a pointer to a pointer to a pointer to a char. So
your first malloc() should create the space for several 'pointer
to a pointer to a char' and make ptr (which is a pointer to a
"pointer to a pointer to char" point to it through the malloc call)

char *** ptr=malloc(MAT * sizeof **ptr);
Wrong. OP was correct. **ptr has type pointer-to-char, and you need
sizeof pointer-to-pointer-to-char.

Yeh, I meant to use the TYPE not the object here so it would be:

char *** ptr=malloc(MAT * sizeof (char**));

Hence the two '**' which was wrong when I used them with sizeof **ptr...

This is /EXACTLY/ why we here in c.l.c don't try to foist the
ridiculous "sizeof (char **)" nonsense on newbies (or even oldbies,
for that matter). If you stick to "sizeof *foo" for every malloc
call you ever write, you will /never/ commit such a serious blunder,
and the regulars won't have to waste time correcting you.

You can think of it more clearly like this:

char array[MAT][ROW][COL];
char * array[MAT][ROW]
char ** array[MAT]
char *** array;
Maybe *you* can; but an experienced C programmer might look at
that list of declarations and see immediately that they're all of
different types, stored and accessed by different methods; and then
he might accuse you of either oversimplification or ignorance -- and
we wouldn't want that, now would we?

In practice the sizeof ***ptr and sizeof ** ptr etc will all be
the same.

Absolute nonsense, particularly given OP's declaration of ptr. Don't
bandy about such dangerous and incorrect advice.

Again I meant sizeof(char ***) and sizeof(char **) would be the same in
practice ...

Still wrong, as far as C is concerned. (char ***) and (char **) are
not compatible types, nor do they necessarily have the same size.
I think my brain suffered a temporary blue screen of death...

I think you're still working through the stage that comes right after
wide-eyed newbieness, which is 1337|\|355. ("Don't assume that just
because you can kick the TV and make it work, that you are an
uber-TV-repairman.") Keep reading c.l.c, though, and you'll pull
through. ;-)

-Arthur

[Arthur J. O'Dwyer]

On Fri, 30 Jan 2004, Vijay Kumar R Zanvar wrote:

Is this a correct method of allocating a 3-dimensional
array?

/* assume all malloc's succeed */
No! Definitely not! ;-D

#define MAT 2
#define ROW 2
#define COL 2

char ***ptr = malloc ( MAT * sizeof *ptr );

for ( i = 0; i < MAT; i++ )
ptr[i] = malloc ( ROW * sizeof **ptr );
I would stick to the c.l.c idiom here, and write

ptr[i] = malloc(ROW * sizeof *ptr[i]);

just to avoid any lingering doubts about how many asterisks
are supposed to go there.
for ( i = 0; i < MAT; i++ )
for ( j = 0; j < ROW; j++ )
ptr[i][j] = malloc ( COL * sizeof ***ptr );

Ditto on this malloc. Except that in practice, I'd fold
these two loops into one, making the finished product look
like this:

char ***new_array_without_error_checking(int MAT, int ROW, int COL)
{
char ***ptr;
int i, j;

ptr = malloc(MAT * sizeof *ptr);
for (i=0; i < MAT; ++i) {
ptr[i] = malloc(ROW * sizeof *ptr[i]);
for (j=0; j < ROW; ++j)
ptr[i][j] = malloc(COL * sizeof *ptr[i][j]);
}

return ptr;
}
Incidentally, I see that the FAQ (q6.16) uses malloc casting,
which it actually POINTS OUT reaches a point where "the syntax
starts getting horrific," so I don't know why Steve Summit added
it in the first place! :-(

HTH,
-Arthur

[E. Robert Tisdale]

Arthur J. O'Dwyer wrote:

Incidentally, I see that the FAQ (q6.16) uses malloc casting,
which it actually POINTS OUT reaches a point
where "the syntax starts getting horrific," so
I don't know why Steve Summit added it in the first place! :-(

Snicker, snicker. Chortle, chortle. Gloat, gloat.

[Sean Kenwrick]

"Arthur J. O'Dwyer" <aj*@nospam.andrew.cmu.edu> wrote in message
news:Pi**********************************@unix43.a ndrew.cmu.edu...

On Fri, 30 Jan 2004, Sean Kenwrick wrote:

"nrk" <ra*********@devnull.verizon.net> wrote...

Sean Kenwrick wrote:
>
> You define a pointer to a pointer to a pointer to a char. So
> your first malloc() should create the space for several 'pointer
> to a pointer to a char' and make ptr (which is a pointer to a
> "pointer to a pointer to char" point to it through the malloc call)
>
> char *** ptr=malloc(MAT * sizeof **ptr);

Wrong. OP was correct. **ptr has type pointer-to-char, and you need
sizeof pointer-to-pointer-to-char.

Yeh, I meant to use the TYPE not the object here so it would be:

char *** ptr=malloc(MAT * sizeof (char**));

Hence the two '**' which was wrong when I used them with sizeof **ptr...

This is /EXACTLY/ why we here in c.l.c don't try to foist the
ridiculous "sizeof (char **)" nonsense on newbies (or even oldbies,
for that matter). If you stick to "sizeof *foo" for every malloc
call you ever write, you will /never/ commit such a serious blunder,
and the regulars won't have to waste time correcting you.

This reason I got into trouble here is /EXACTLY/ because I used sizeof
(*foo) instead to sizeof(type). When you see a statement like sizeof
***foo or sizeof **foo you cannot immediately descern from the code what is
going on without hunting back to track down the declaration of foo then
figuring out it's indirection so that you can know what ***foo is refering
to (is it the object, a pointer to the object or a pointer to a pointer ot
the object??? (horrible!)). Whereas with sizeof (char**) is immediately
clear what it is refering to.

The only advantage I can see of using sizeof *foo is that it protects you
from the case where you might change the fundamental type of foo in the
future, but this is a very rare occurance for something that destroys the
readability of your code...

I'm sure you will enlighten me as to othjer reasons why you prefer sizeof
*foo here in c.l.c and I might yet be converted if you present a good enough
case :-)

Sean

[Richard Heathfield]

Sean Kenwrick wrote:

This reason I got into trouble here is /EXACTLY/ because I used sizeof
(*foo) instead to sizeof(type).
Er, no, the reason you got into trouble was that you used sizeof **foo
instead of sizeof *foo.
When you see a statement like sizeof
***foo or sizeof **foo you cannot immediately descern from the code what
is going on without hunting back to track down the declaration of foo then
figuring out it's indirection so that you can know what ***foo is refering
to (is it the object, a pointer to the object or a pointer to a pointer ot
the object??? (horrible!)).
It's really, really easy. Whatever it is that's on the left of the = malloc,
you do sizeof *whatever. Watch:

T *p = malloc(n * sizeof *p); /* generic example */

char *******************s = malloc(n * sizeof *s); /* easy, huh? */

if(s == NULL) abort();

for(i = 0; i < n; i++)
{
s[i] = malloc(m * sizeof *s[i]);
if(s[i] == NULL) abort();
}

Here, we see a possible way in which you can get tangled if you aren't
simple-minded enough. The basic algorithm here is:

1) write the pointer down - in this case, s[i]
2) write = malloc(numobjects * sizeof *
3) copy the value you wrote in step 1, and paste it at the end of the
line.
4) add );

Let's demonstrate by pointing this pointer: d[e].m[17]->o

at some allocated space.

Step 1: we write the pointer down:

d[e].m[17]->o

Step 2: Now we put " = malloc(numobjects * sizeof *", which gives us:

d[e].m[17]->o = malloc(numobjects * sizeof *

Step 3: We copy/paste the original pointer:

d[e].m[17]->o = malloc(numobjects * sizeof *d[e].m[17]->o

Step 4: make good:

d[e].m[17]->o = malloc(numobjects * sizeof *d[e].m[17]->o);
Whereas with sizeof (char**) is immediately
clear what it is refering to.

Please verify that this call is correct, using only information that is
immediately clear:

d[e].m[17]->o = malloc(numobjects * sizeof(int(*)(int, char **)));

Can you verify that this code is correct? I certainly can't.

--
Richard Heathfield : bi****@eton.powernet.co.uk
"Usenet is a strange place." - Dennis M Ritchie, 29 July 1999.
C FAQ: http://www.eskimo.com/~scs/C-faq/top.html
K&R answers, C books, etc: http://users.powernet.co.uk/eton

[Amos Leffler]

Richard Heathfield wrote:

Sean Kenwrick wrote:

This reason I got into trouble here is /EXACTLY/ because I used sizeof
(*foo) instead to sizeof(type).

Er, no, the reason you got into trouble was that you used sizeof **foo
instead of sizeof *foo.

When you see a statement like sizeof
***foo or sizeof **foo you cannot immediately descern from the code what
is going on without hunting back to track down the declaration of foo then
figuring out it's indirection so that you can know what ***foo is refering
to (is it the object, a pointer to the object or a pointer to a pointer ot
the object??? (horrible!)).

It's really, really easy. Whatever it is that's on the left of the = malloc,
you do sizeof *whatever. Watch:

T *p = malloc(n * sizeof *p); /* generic example */

char *******************s = malloc(n * sizeof *s); /* easy, huh? */

if(s == NULL) abort();

for(i = 0; i < n; i++)
{
s[i] = malloc(m * sizeof *s[i]);
if(s[i] == NULL) abort();
}

Here, we see a possible way in which you can get tangled if you aren't
simple-minded enough. The basic algorithm here is:

1) write the pointer down - in this case, s[i]
2) write = malloc(numobjects * sizeof *
3) copy the value you wrote in step 1, and paste it at the end of the
line.
4) add );

Let's demonstrate by pointing this pointer: d[e].m[17]->o

at some allocated space.

Step 1: we write the pointer down:

d[e].m[17]->o

Step 2: Now we put " = malloc(numobjects * sizeof *", which gives us:

d[e].m[17]->o = malloc(numobjects * sizeof *

Step 3: We copy/paste the original pointer:

d[e].m[17]->o = malloc(numobjects * sizeof *d[e].m[17]->o

Step 4: make good:

d[e].m[17]->o = malloc(numobjects * sizeof *d[e].m[17]->o);

Whereas with sizeof (char**) is immediately
clear what it is refering to.

Please verify that this call is correct, using only information that is
immediately clear:

d[e].m[17]->o = malloc(numobjects * sizeof(int(*)(int, char **)));

Can you verify that this code is correct? I certainly can't.

--
Richard Heathfield : bi****@eton.powernet.co.uk
"Usenet is a strange place." - Dennis M Ritchie, 29 July 1999.
C FAQ: http://www.eskimo.com/~scs/C-faq/top.html
K&R answers, C books, etc: http://users.powernet.co.uk/eton

Hi,
If anybody is interested there is a discussion of this problem in
"Numerical Recipes in C", 2nd edn. It covers two dimensional arrays but
can be extended if you are interested.
Amos L.

[Sean Kenwrick]

"Richard Heathfield" <do******@address.co.uk.invalid> wrote in message
news:bv**********@hercules.btinternet.com...

Sean Kenwrick wrote:

This reason I got into trouble here is /EXACTLY/ because I used sizeof
(*foo) instead to sizeof(type).
Er, no, the reason you got into trouble was that you used sizeof **foo
instead of sizeof *foo.

When you see a statement like sizeof
***foo or sizeof **foo you cannot immediately descern from the code what
is going on without hunting back to track down the declaration of foo then figuring out it's indirection so that you can know what ***foo is refering to (is it the object, a pointer to the object or a pointer to a pointer ot the object??? (horrible!)).

It's really, really easy. Whatever it is that's on the left of the =

malloc, you do sizeof *whatever. Watch:

T *p = malloc(n * sizeof *p); /* generic example */

char *******************s = malloc(n * sizeof *s); /* easy, huh? */

if(s == NULL) abort();

for(i = 0; i < n; i++)
{
s[i] = malloc(m * sizeof *s[i]);
if(s[i] == NULL) abort();
}

Here, we see a possible way in which you can get tangled if you aren't
simple-minded enough. The basic algorithm here is:

1) write the pointer down - in this case, s[i]
2) write = malloc(numobjects * sizeof *
3) copy the value you wrote in step 1, and paste it at the end of the
line.
4) add );

Let's demonstrate by pointing this pointer: d[e].m[17]->o

at some allocated space.

Step 1: we write the pointer down:

d[e].m[17]->o

Step 2: Now we put " = malloc(numobjects * sizeof *", which gives us:

d[e].m[17]->o = malloc(numobjects * sizeof *

Step 3: We copy/paste the original pointer:

d[e].m[17]->o = malloc(numobjects * sizeof *d[e].m[17]->o

Step 4: make good:

d[e].m[17]->o = malloc(numobjects * sizeof *d[e].m[17]->o);

OK I'm sold!

The OP had code like this:

for ( i = 0; i < MAT; i++ )
for ( j = 0; j < ROW; j++ )
ptr[i][j] = malloc ( COL * sizeof ***ptr );
Which I was just going to use as a counter example to your point until I
realised that ptr[i][j] dereferences to **ptr, so that malloc(COL*sizeof
***ptr) (following your rules) is correct without needing to know the
original type of ptr.

Sean

[pete]

Sean Kenwrick wrote:

"Richard Heathfield" <do******@address.co.uk.invalid> wrote in message
news:bv**********@hercules.btinternet.com...

It's really, really easy.
Whatever it is that's on the left of the = malloc,
you do sizeof *whatever. Watch:

T *p = malloc(n * sizeof *p); /* generic example */

char *******************s = malloc(n * sizeof *s);
/* easy, huh? */

if(s == NULL) abort();

for(i = 0; i < n; i++)
{
s[i] = malloc(m * sizeof *s[i]);
if(s[i] == NULL) abort();
}

Here, we see a possible way in which you can get
tangled if you aren't
simple-minded enough. The basic algorithm here is:

1) write the pointer down - in this case, s[i]
2) write = malloc(numobjects * sizeof *
3) copy the value you wrote in step 1,
and paste it at the end of the line.
4) add );

Let's demonstrate by pointing this pointer: d[e].m[17]->o

at some allocated space.

Step 1: we write the pointer down:

d[e].m[17]->o

Step 2: Now we put " = malloc(numobjects * sizeof *",
which gives us:

d[e].m[17]->o = malloc(numobjects * sizeof *

Step 3: We copy/paste the original pointer:

d[e].m[17]->o = malloc(numobjects * sizeof *d[e].m[17]->o

Step 4: make good:

d[e].m[17]->o = malloc(numobjects * sizeof *d[e].m[17]->o);

OK I'm sold!

It's good even if the left operand of the assignment has
side effects, because sizeof doesn't evaluate it's operand.

I like the idiom for allocations for arrays of all types,
except when allocating bytes for strings.
Then I like malloc(length + 1).

--
pete

[Old Wolf]

> Is this a correct method of allocating a 3-dimensional

array?

/* assume all malloc's succeed */
#define MAT 2
#define ROW 2
#define COL 2

char ***ptr = malloc ( MAT * sizeof *ptr );

for ( i = 0; i < MAT; i++ )
ptr[i] = malloc ( ROW * sizeof **ptr );

for ( i = 0; i < MAT; i++ )
for ( j = 0; j < ROW; j++ )
ptr[i][j] = malloc ( COL * sizeof ***ptr );

[note - code is even more voluminous when you include all the malloc
NULL checks and free() calls]

How is this any better than:
char (*ptr)[ROW][COL] = malloc(MAT * sizeof *ptr);

If your answer is that MAT, ROW and COL might not be compiletime
constants after all, then you can go:
char *ptr = malloc(MAT * ROW * COL * sizeof *ptr)
and
#define CELL(mat,row,col) ptr[(mat)*ROW*COL + (row)*COL + (col)]
or similar

If your answer is re. row-major vs. col-major order, then you can
do a little syntax rearranging to fix that.

[Arthur J. O'Dwyer]

On Mon, 2 Feb 2004, Sean Kenwrick wrote:

"Richard Heathfield" <do******@address.co.uk.invalid> wrote...

Sean Kenwrick wrote:

This reason I got into trouble here is /EXACTLY/ because I used sizeof
(*foo) instead to sizeof(type).
Er, no, the reason you got into trouble was that you used sizeof **foo
instead of sizeof *foo. It's really, really easy. Whatever it is that's on the left of the
= malloc, you do sizeof *whatever. Watch: Let's demonstrate by pointing this pointer: d[e].m[17]->o
at some allocated space. d[e].m[17]->o = malloc(numobjects * sizeof *d[e].m[17]->o);

OK I'm sold!

The OP had code like this:

for ( i = 0; i < MAT; i++ )
for ( j = 0; j < ROW; j++ )
ptr[i][j] = malloc ( COL * sizeof ***ptr );
Which I was just going to use as a counter example to your point until I
realised that ptr[i][j] dereferences to **ptr,

That is meaningless. ptr[i][j] is the j'th element of the array
whose first element is pointed to by the pointer expression ptr[i].
**ptr is the object pointed to by the pointer expression *ptr.
The two are very rarely the same object, although they do have the
same size.
so that malloc(COL*sizeof ***ptr) (following your rules) is correct

That it is, but I suspect most C programmers here would be even
more comfortable if you had followed the simple "four-step" rule
that Richard just told you, right above: write the pointer, write
"= malloc(COL * sizeof *", write the pointer again, and then write
");". This foolproof method yields

ptr[i][j] = malloc(COL * sizeof *ptr[i][j]);

which really /is/ obviously correct, even to those lazy bums who
don't want to waste precious debugging time comparing the numbers of
asterisks and square brackets!
[See my post farther up in this thread, which uses the "four-step"
method exclusively.]

-Arthur

[Richard Heathfield]

Arthur J. O'Dwyer wrote:

On Mon, 2 Feb 2004, Sean Kenwrick wrote:

so that malloc(COL*sizeof ***ptr) (following your rules) is correct

That it is, but I suspect most C programmers here would be even
more comfortable if you had followed the simple "four-step" rule
that Richard just told you, right above: write the pointer, write
"= malloc(COL * sizeof *", write the pointer again, and then write
");". This foolproof method yields

ptr[i][j] = malloc(COL * sizeof *ptr[i][j]);

which really /is/ obviously correct, even to those lazy bums who
don't want to waste precious debugging time comparing the numbers of
asterisks and square brackets!
[See my post farther up in this thread, which uses the "four-step"
method exclusively.]

"The four-step method" sounds terribly complicated, doesn't it?
Coming soon: how to write "Hello, world." on the standard output device in
only 85 steps!

Preview:

Step 1: Press #
Step 2: Press i
Step 3: Press n
Step 4: Press c

...

:-)

--
Richard Heathfield : bi****@eton.powernet.co.uk
"Usenet is a strange place." - Dennis M Ritchie, 29 July 1999.
C FAQ: http://www.eskimo.com/~scs/C-faq/top.html
K&R answers, C books, etc: http://users.powernet.co.uk/eton

[Richard Heathfield]

Amos Leffler wrote:

Hi,
If anybody is interested there is a discussion of this problem in
"Numerical Recipes in C", 2nd edn. It covers two dimensional arrays but
can be extended if you are interested.

The C code in Numerical Recipes is a disaster. They've gone to huge lengths
to keep the 1-based arrays from their Fortran code. This was a mistake
which permeates almost all their code. If you were looking for undefined
behaviour, look no further!

--
Richard Heathfield : bi****@eton.powernet.co.uk
"Usenet is a strange place." - Dennis M Ritchie, 29 July 1999.
C FAQ: http://www.eskimo.com/~scs/C-faq/top.html
K&R answers, C books, etc: http://users.powernet.co.uk/eton

[Dave Thompson]

On Fri, 30 Jan 2004 17:27:57 -0500 (EST), "Arthur J. O'Dwyer"
<aj*@nospam.andrew.cmu.edu> wrote:
<snip>

Incidentally, I see that the FAQ (q6.16) uses malloc casting,
which it actually POINTS OUT reaches a point where "the syntax
starts getting horrific," so I don't know why Steve Summit added
it in the first place! :-(

It was there since pre-C89, when it was needed -- and in practice was
often needed at least for portability for several years after 89. It
was removed in Feb 1999 (there was discussion at the time) from the
text version (periodically posted, rtfm and I believe faqs.org) but
Steve has never (yet AFAIK!) gotten that version into the website.

- David.Thompson1 at worldnet.att.net