freeing pointers in pointer array causes SegFault [solved]

[Jordanwb]

Hello. I have the following code block:

#include <stdlib.h>

typedef struct List_s
{
int count;
void **items;
} List;

List *list_create()
{
List *list = malloc (sizeof (List));
list->count = 0;
list->items = malloc (1 * sizeof (void *));

return list;

}

void list_add (List *list, void *item)
{
list->items = realloc (list->items, (list->count + 1) * sizeof (void *));
list->items[list->count] = item;
list->count++;
}

void list_dealloc (List *list)
{
for (int i = 0; i < list->count; i++)
{
free (list->items[i]);
}

free (list->items);
free (list);

}

When I pass a List to list_dealloc I get a segmentation fault on the line in the for loop. I can print the contents of items[i] just fine (in my test case it's the string "Hello"). I'm using gcc-4.5.1 on 64 bit Fedora 14 if that's anything useful.

modified on Monday, February 28, 2011 12:57 PM

[Cool_Dev]

Jordanwb wrote:

segmentation fault on the line in the for loop

If the crash is at free() function, ensure that you are passing a heap pointer as item to list_add() function, not the address of a stack variable.

[Lost User]

Are you sure that all the items passed in to your list through the list_add() function have been created by malloc()?

I must get a clever new signature for 2011.

[Jordanwb]

Thanks for replying. This is what's in my main() function:

List *list = list_create();
list_add (list, "Hello");
list_dealloc (list);

[Lost User]

So you are adding a pointer to a constant string which you later try to free(): result SEGV fault. You need to ensure that every pointer you pass into list_add() is pointing to a memory block that has been returned from malloc(). something like:

List *list = list_create();
char* pitem = (char*)malloc(10);
strcpy(pitem, "Hello");
list_add (list, pitem);
// pitem may now be removed with a call to free()
list_dealloc (list);

I must get a clever new signature for 2011.

[Jordanwb]

Okay thanks.