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sizeof dynamic array

 
 
D@nny
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Posts: n/a
 
      11-07-2004
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.
 
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Mike Wahler
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      11-07-2004
"D@nny" <(E-Mail Removed)> wrote in message
news:fZljd.1744$(E-Mail Removed)...
> 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


 
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Martin Ambuhl
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      11-07-2004
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.
 
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Joona I Palaste
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      11-07-2004
D@nny <(E-Mail Removed)> 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 ((E-Mail Removed)) ------------- Finland --------\
\-------------------------------------------------------- rules! --------/
"To know me IS to love me."
- JIPsoft
 
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Al Bowers
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      11-07-2004


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: http://www.velocityreviews.com/forums/(E-Mail Removed) (remove the x to send email)
http://www.geocities.com/abowers822/

 
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Skybuck Flying
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Posts: n/a
 
      11-07-2004

"Mike Wahler" <(E-Mail Removed)> wrote in message
news:cfmjd.7863$(E-Mail Removed) ink.net...
> "D@nny" <(E-Mail Removed)> wrote in message
> news:fZljd.1744$(E-Mail Removed)...
> > 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.


 
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Barry Schwarz
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Posts: n/a
 
      11-07-2004
On Sun, 7 Nov 2004 15:47:26 +0100, "Skybuck Flying"
<(E-Mail Removed)> wrote:

>
>"Mike Wahler" <(E-Mail Removed)> wrote in message
>news:cfmjd.7863$(E-Mail Removed) 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>>
 
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Method Man
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Posts: n/a
 
      11-08-2004

"D@nny" <(E-Mail Removed)> wrote in message
news:fZljd.1744$(E-Mail Removed)...
> 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.


 
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Default User
Guest
Posts: n/a
 
      11-08-2004

Method Man wrote:
> "D@nny" <(E-Mail Removed)> wrote in message
> news:fZljd.1744$(E-Mail Removed)...
> > 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

 
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Mike Wahler
Guest
Posts: n/a
 
      11-09-2004

"Method Man" <(E-Mail Removed)> wrote in message
news:H1Ljd.5882$(E-Mail Removed)...
>
> "D@nny" <(E-Mail Removed)> wrote in message
> news:fZljd.1744$(E-Mail Removed)...
> > 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


 
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