sizeof dynamic array

[D]

hi,

i would like to know how to calculate the size of a dynamic array
created using a dereference declaration like int *numbers and
allocating via malloc or calloc: numbers=(int
*)malloc(sizeof(int)*10);

using sizeof(numbers) will return the pointers size...

is there a possibility?

D@nny
--
life already is expensive, so why waste money on expensive
software.

[Mike Wahler]

"D@nny" <[e-mail]> wrote in message
news:fZ****************@amsnews03-serv.chello.com...

hi,

i would like to know how to calculate the size of a dynamic array
Determine how big you need it to be, and allocate that much.
created using a dereference declaration like int *numbers
That's not a 'dereference declaration'. Its a declaration
of a pointer object.
and
allocating via malloc or calloc: numbers=(int
*)malloc(sizeof(int)*10);
Never cast the return value from 'malloc()'. See the FAQ
for details.

using sizeof(numbers) will return the pointers size...
Yes. Because 'numbers' is a pointer.

is there a possibility?

Yes. You know the size when you allocate it. Just retain
that value (i.e. store it in a 'size_t' object).

int *numbers = 0;
size_t sz = 10 * sizeof *numbers;

if(numbers = malloc(sz))
{
printf("%lu bytes allocated.\n", (unsigned long)sz);
/* do stuff */
free(numbers);
}

-Mike

[Martin Ambuhl]

D@nny wrote:

hi,

i would like to know how to calculate the size of a dynamic array
created using a dereference declaration like int *numbers and
allocating via malloc or calloc: numbers=(int *)malloc(sizeof(int)*10);
Guess what: the size of the array is sizeof(int)*10. What did you think
it was.

BTW, your code line above is not, at least here, idiomatic. Had you
lurked before posting (expected usenet behavior) or checked the FAQ
before posting (expected usenet behavior), you would have no this.

You have no idea how many people everyday violate those two standard
rules every day just to post code that casts the return value from
malloc (corrected several times a day here), uses hard-coded magic
numbers, uses hard-coded data types in allocation statements, asks
questions answered multiple times a week (as is yours), and fails to
include a compilable example.

#include <stdlib.h>
int main(void)
{
const number_elements = 10;
int *numbers;
numbers = malloc(number_elements * sizeof *numbers);
if (!numbers) { /* handle error */ }
else free(numbers);
return 0;
}
using sizeof(numbers) will return the pointers size...

is there a possibility?

Obviously, there is. If you can figure out how much space to ask malloc
for, you already know the answer.

[Joona I Palaste]

D@nny <[e-mail]> scribbled the following:

hi, i would like to know how to calculate the size of a dynamic array
created using a dereference declaration like int *numbers and
allocating via malloc or calloc: numbers=(int
*)malloc(sizeof(int)*10); using sizeof(numbers) will return the pointers size... is there a possibility?

Not in standard C. You'll have to use implementation-specific tricks.
But since you know the size when you allocate the memory, why not simply
keep track of it?

--
/-- Joona Palaste (pa*****@cc.helsinki.fi) ------------- Finland --------\
\-------------------------------------------------------- rules! --------/
"To know me IS to love me."
- JIPsoft

[Al Bowers]

D@nny wrote:

hi,

i would like to know how to calculate the size of a dynamic array
created using a dereference declaration like int *numbers and
allocating via malloc or calloc: numbers=(int
*)malloc(sizeof(int)*10);

using sizeof(numbers) will return the pointers size...

is there a possibility?

What you should do is declare a variable, ex. size_t nelements,
that you can use to store the number of elements allocated for
the int array. If you dynamically allocated the array for
10 elements, then store in nelements the value 10. If you
modify the number of array elements, then modify nelements
appropriately. In the occasion where you might need the
total size of the allocation, just multiply the sizeof(int)
with nelements.

#include <stdio.h>
#include <stdlib.h>

int main(void)
{
int *iarr, *tmp;
size_t nelements = 0;

iarr = malloc((sizeof *iarr)*10);
if(iarr) nelements = 10;
printf("Number of elements allocated is %u\n"
"Size of the allocation is %u\n\n",
nelements, nelements*sizeof(int));
if(nelements)
{
puts("An attempt to increase the allocation one element\n");
tmp = realloc(iarr,(sizeof *iarr)*(nelements+1));
if(tmp)
{
iarr = tmp;
nelements++;
}
printf("Number of elements allocated is %u\n"
"Size of the allocation is %u\n\n",
nelements, nelements*sizeof(int));
}
free(iarr);
return 0;
}
--
Al Bowers
Tampa, Fl USA
mailto: xa******@myrapidsys.com (remove the x to send email)
http://www.geocities.com/abowers822/

[Skybuck Flying]

"Mike Wahler" <mk******@mkwahler.net> wrote in message
news:cf*****************@newsread3.news.pas.earthl ink.net...

"D@nny" <[e-mail]> wrote in message
news:fZ****************@amsnews03-serv.chello.com...

hi,

i would like to know how to calculate the size of a dynamic array

Determine how big you need it to be, and allocate that much.

created using a dereference declaration like int *numbers

That's not a 'dereference declaration'. Its a declaration
of a pointer object.

and
allocating via malloc or calloc: numbers=(int
*)malloc(sizeof(int)*10);

Never cast the return value from 'malloc()'. See the FAQ
for details.

Hello I am not the original poster... but this seems interesting.

I can't the answer in this FAQ;

http://www.eskimo.com/~scs/C-faq/s7.html

Why shouldn't it be typecasted ?

1. Maybe because it doesn't compile on true c compilers ?

2. Or does it lead to bugs ?

Bye,
Skybuck.

[Barry Schwarz]

On Sun, 7 Nov 2004 15:47:26 +0100, "Skybuck Flying"
<no****@hotmail.com> wrote:

"Mike Wahler" <mk******@mkwahler.net> wrote in message
news:cf*****************@newsread3.news.pas.earth link.net...

snip

Never cast the return value from 'malloc()'. See the FAQ
for details.

Hello I am not the original poster... but this seems interesting.

I can't the answer in this FAQ;

http://www.eskimo.com/~scs/C-faq/s7.html

Why shouldn't it be typecasted ?

a. It is unnecessary in C. Superfluous casts should always be
avoided.

b. It leads to undefined behavior (under C89, currently the standard
implemented by most popular compilers) if the prototype for malloc is
omitted.
<<Remove the del for email>>

[Method Man]

"D@nny" <[e-mail]> wrote in message
news:fZ****************@amsnews03-serv.chello.com...

hi,

i would like to know how to calculate the size of a dynamic array
created using a dereference declaration like int *numbers and
allocating via malloc or calloc: numbers=(int
*)malloc(sizeof(int)*10);

using sizeof(numbers) will return the pointers size...

is there a possibility?

As others have mentioned, keep track of the size when you allocate the
array.

I suppose one possible exception is a null-terminated char array where you
can call 'strlen' to get the length of the array.

[Default User]

Method Man wrote:

"D@nny" <[e-mail]> wrote in message
news:fZ****************@amsnews03-serv.chello.com...

hi,

i would like to know how to calculate the size of a dynamic array

As others have mentioned, keep track of the size when you allocate the array.

I suppose one possible exception is a null-terminated char array where you can call 'strlen' to get the length of the array.

That gives you length of the string, not the size of the array. They
are often different. So for instance, you have no way of knowing
whether you can extend the string safely unless you know the size of
the underlying array.

Brian

[Mike Wahler]

"Method Man" <a@b.c> wrote in message
news:H1*******************@read2.cgocable.net...

"D@nny" <[e-mail]> wrote in message
news:fZ****************@amsnews03-serv.chello.com...

hi,

i would like to know how to calculate the size of a dynamic array
created using a dereference declaration like int *numbers and
allocating via malloc or calloc: numbers=(int
*)malloc(sizeof(int)*10);

using sizeof(numbers) will return the pointers size...

is there a possibility?

As others have mentioned, keep track of the size when you allocate the
array.

I suppose one possible exception is a null-terminated char array where you
can call 'strlen' to get the length of the array.

char *array = malloc(100);
size_t sz = 0;

if(array)
{
strcpy(array, "Hello");
sz = strlen(array); /* sz != 100 */
}

free(array);

-Mike

[Method Man]

"Mike Wahler" <mk******@mkwahler.net> wrote in message
news:mk******************@newsread1.news.pas.earth link.net...

"Method Man" <a@b.c> wrote in message
news:H1*******************@read2.cgocable.net...

"D@nny" <[e-mail]> wrote in message
news:fZ****************@amsnews03-serv.chello.com...

hi,

i would like to know how to calculate the size of a dynamic array
created using a dereference declaration like int *numbers and
allocating via malloc or calloc: numbers=(int
*)malloc(sizeof(int)*10);

using sizeof(numbers) will return the pointers size...

is there a possibility?

As others have mentioned, keep track of the size when you allocate the
array.

I suppose one possible exception is a null-terminated char array where you can call 'strlen' to get the length of the array.

char *array = malloc(100);
size_t sz = 0;

if(array)
{
strcpy(array, "Hello");
sz = strlen(array); /* sz != 100 */
}

free(array);

Yeah, I should have noted that 'strlen' gives the length of the string and
not necessarily the allocated size of the array. That's why I said "possible
exception".

[Mike Wahler]

"Method Man" <a@b.c> wrote in message
news:cw*******************@read2.cgocable.net...

"Mike Wahler" <mk******@mkwahler.net> wrote in message
news:mk******************@newsread1.news.pas.earth link.net...

"Method Man" <a@b.c> wrote in message
news:H1*******************@read2.cgocable.net...

"D@nny" <[e-mail]> wrote in message
news:fZ****************@amsnews03-serv.chello.com...
> hi,
>
> i would like to know how to calculate the size of a dynamic array
> created using a dereference declaration like int *numbers and
> allocating via malloc or calloc: numbers=(int
> *)malloc(sizeof(int)*10);
>
> using sizeof(numbers) will return the pointers size...
>
> is there a possibility?
>

As others have mentioned, keep track of the size when you allocate the
array.

I suppose one possible exception is a null-terminated char array where you can call 'strlen' to get the length of the array.
char *array = malloc(100);
size_t sz = 0;

if(array)
{
strcpy(array, "Hello");
sz = strlen(array); /* sz != 100 */
}

free(array);

Yeah, I should have noted that 'strlen' gives the length of the string and
not necessarily the allocated size of the array.

Right. The point is that there is no way to determine after the fact
how much memory was allocated. Another 'subtle' point: 'malloc()'
is required to attempt to allocate *at least* the requested size.
It is allowed to (and often does) allocate more (but the program
must only access within bounds of the requested size).
That's why I said "possible
exception".

In standard C, no exceptions.

-Mike