total size of dynamic memory allocated array

hi, this is my first post and my question is...

When using the sizeof operator on an array name it returns
the size that the whole array allocates,eg

Code:

---

int array[ 20 ] = { 0 };
printf( "%u ", sizeof( array ) ); /* returns 80 */

---

is there a way to do the same on a dynamic array? eg

Code:

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int *array = ( int * ) malloc( 20 * sizeof( int ) );

---

thanks

Of course there is, you can dynamically allocate an array with only a little bit more work than a static array. But if you try to use the sizeof operator on the entire dynamic array then the value will most likely be 4, which is the size of a pointer. sizeof is determined at compile time and the size of a dynamic array is determined at runtime, so sizeof will only return the size of the pointer and not the memory allocated.

Code:

---

/* pseudocode */
#include <stdio.h>
#include <stdlib.h>

int main ( void )
{
  int i, errors = 0;
  char **array = malloc ( 20 * sizeof ( char * ) );
  if ( array != NULL ) {
    for ( i = 0; i < 20; i++ ) {
      array[i] = malloc ( 10 * sizeof ( char ) );
      if ( array[i] == NULL ) errors++;
    }
    for ( i = 0; i < 20; i++ )
      free ( array[i] );
    free ( array );
  }
  printf ( "Code exited with %d errors\n", errors );
  return EXIT_SUCCESS;
}

---

-Prelude

well, i might be stupid :confused: but i can't see how your code answers my question. So, i'll try to be more specific.

After i allocate memory for my pointer i pass him as an arg
to an other function. I want that function to be able to get the size of memory allocation...

i hope i made myself clear,
thanks again

>but i can't see how your code answers my question
I thought you were asking both how to create a dynamic array and how to get the size of it. The code doesn't answer the latter, but the message does:

Quote:

---

But if you try to use the sizeof operator on the entire dynamic array then the value will most likely be 4, which is the size of a pointer. sizeof is determined at compile time and the size of a dynamic array is determined at runtime, so sizeof will only return the size of the pointer and not the memory allocated.

---

-Prelude

thanks Prelude...
here's what i want to do ( kind of... )

Code:

---

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

void f( int *ptr )
{
        int *innerPtr, size = 0;

        /* here i want to count the allocated memory of ptr
        and assign it in variable size */
       
        innerPtr = ( int * ) malloc( size * sizeof( int ) );

        /* rest of the code */

        return;
}

int main( void )
{
        int *array;

        array = ( int * ) malloc( 20 * sizeof( int ) );

        f( array );

        return 0;
}

---

thanks for any additional input

Set the size variable in main and pass it to the function. That way tends to be less buggy than trying to acquire the size of a dynamic array.

Code:

---

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

static void f( int *ptr, int size )
{
  int *innerPtr;       
  innerPtr = malloc( size * sizeof *innerPtr );
  /* rest of the code */
  return;
}

int main( void )
{
  int *array, size = 20;
  array = malloc( size * sizeof *array );
  f( array, size );
  return 0;
}
/* Don't forget to free the pointers :) */

---

-Prelude

isn't there a getallocsize function or something like that?
i've been searching all afternoon...

>isn't there a getallocsize function or something like that?
Nothing that's standard

-Prelude

> After i allocate memory for my pointer i pass him
> as an arg to an other function. I want that function
> to be able to get the size of memory allocation...

No, there is no such "standard" function. You, however, can write one if you like--

All that is required is that you learn enough about Window's or DOS's memory management so that you can break apart the block header for the dynamically allocated RAMblock in questions.

Normally, an allocated block has about 12 bytes extra prepended to it. This additional space (that you normally aren't aware of), describes information about the block (lockstate, size, offset-size, etc.).

enjoy.

Or as an alternative you could wrap your pointer in a struct, along with a variable containing the size.

well i'm interested in low-level programming and i would like
to know if there is any kind of book that could help me do so.
The unusual thing about me is that in uni i use visual studio
on pc and back home i program on a macintosh with
codewarrior. However info on windows memory management
won't hurt :)

tia

Quote:

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Originally posted by Sayeh
> After i allocate memory for my pointer i pass him
> as an arg to an other function. I want that function
> to be able to get the size of memory allocation...

No, there is no such "standard" function. You, however, can write one if you like--

All that is required is that you learn enough about Window's or DOS's memory management so that you can break apart the block header for the dynamically allocated RAMblock in questions.

Normally, an allocated block has about 12 bytes extra prepended to it. This additional space (that you normally aren't aware of), describes information about the block (lockstate, size, offset-size, etc.).

enjoy.

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