realloc and array notation?

Unregistered · 01-05-2002

This is probably a simple question but I'd like a little clarification.
Do these two code pieces do the same thing with the same result?

Code:

int array[2] = {1,2};
int *temp;

temp = realloc(array, 3 * sizeof array);
if(temp != NULL)
    array = temp;
array[2] = 3;
printf("%d", array[2]);

and

Code:

int *array;
int *temp;

array = malloc(2 * sizeof *array);
if(array != NULL){
    array[0] = 1;
    array[1] = 2;
}
temp = realloc(array, 3 * sizeof *array);
if(temp != NULL)
    array = temp;
array[2] = 3;
printf("%d", array[2]);

Both of these pieces should change the size of the array to 3, put the integer 3 into the third element and print it without any access violation errors, right?

**Salem** · 01-05-2002

> Do these two code pieces do the same thing with the same result?
No
The first example is illegal - you can't resize an array

The second example does as you suggest

**Deckard** · 01-05-2002

From the realloc() manual page:
Unless ptr is NULL, it must have been returned by an earlier call to malloc(), calloc() or realloc().

In your case, 'array' is what the man page is referring to as 'ptr'. Its address must have been returned by a previous call to malloc(), calloc(), or realloc().

Also, 'array' is on the stack in the first example, and is a pointer to the heap in the second. If realloc() cannot find the contiguous memory it needs after 'array', it will allocate enough memory elsewhere, copy the contents of 'array' to the new location, and then (try to) free the memory originally used by 'array' (thus trashing the stack in the first example).

Since a pointer (temp) is pushed onto the stack after array, we know there isn't enough room to simply expand the size of 'array', leading to the trashed stack.

The second example should work fine, whereas the first will core dump.

**-KEN-** · 01-05-2002

The first is illegal, and...

whenver you multiply by sizeof *array, you really shouldn't. you're just wasting space...
example:
temp = realloc(array, 3 * sizeof array);

see, an integer is 4 bytes
*array is an integer pointer (4 bytes)
3 x 4 = 12 when all you wanted is 3, right?
so that works out better as:
temp = realloc(array, 3);

same with anywhere else you multiply by sizeof *array.

//edit: wow, I just posted this, and in the time it took me to write this 3 other people had replied when before there was noone! weird....

**Salem** · 01-05-2002

You got them the wrong way round Ken

sizeof array is sizeof int*
sizeof *array is sizeof int - array is dereferenced, and then the size is taken

> temp = realloc(array, 3);
and this is just 3 bytes, not 3 integers

**-KEN-** · 01-05-2002

right...feeling stupid now...

:smacking self: I figured out why I was thinking that, it's because the only time I've ever had to use malloc() or realloc() was with char's....I was just thinking along those lines...

...still feeling stupid...

Unregistered · 01-05-2002

Wow, a lot of replies. So I think enough people have agreed that the first way is wrong and the second is right, I have to create the array with malloc to use realloc and resize it.

After it's allocated I can use array notation to access the data, right?

**Salem** · 01-05-2002

> After it's allocated I can use array notation to access the data, right?
yes you can

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