process heap is organized as chunk of free memory need not be
contiguous and during freeing what free() does is update it's book
keeping records and it's bins array so that next time if asked for new
space may return the same space however it doesn't wipe the freed data
but few run time directive are present in few rtdl libary to overwrite
the freed space with some poison value ( configurable )
On Sun, Nov 13, 2011 at 11:52 PM, sumit mahamuni
<sumit143smail...@gmail.com> wrote:
>
>
> On Nov 11, 9:16 pm, saurabh singh <saurab...@gmail.com> wrote:
>> well that would be tough for the compiler to predict things that will
>> happen during run time.Its the job of garbage collector to do that.
> well compiler cant predict what will happen at runtime.
>>
>>
>>
>>
>>
>>
>>
>>
>>
>> On Fri, Nov 11, 2011 at 8:36 PM, shady <sinv...@gmail.com> wrote:
>> > ok, thanks.
>>
>> > why do we need to free the memory ?
> if you do not free the memory, you will run out of the memory.
>
>> > Suppose i have a linked list of 1000 nodes and i make the head of it =
>> > NULL, thus losing the whole list. Then compiler can look at other variables
>> > and if this list has not been referenced anywhere else then it is useless,
>> > thus will free the memory.
>> > Is the argument wrong ?
>
>>
>> > On Fri, Nov 11, 2011 at 8:20 PM, vikas <vikas.rastogi2...@gmail.com>wrote:
>>
>> >> nopes , they are not connected, it is just a chance you are getting
>> >> the same values and nothing is overwritten there: basically these are
>> >> DANGLING POINTERS . Now you should keep practising something like this
>>
>> >> #define FREE(N) { free(N); N=NULL;}
>>
>> >> to avoid such mistakes
>>
>> >> On Nov 11, 3:41 pm, shady <sinv...@gmail.com> wrote:
>> >> > typedef struct n{
>> >> > int num;
>> >> > struct n *next;
>>
>> >> > }node;
>>
>> >> > node is the structure to create the linked list.
>>
>> >> > node *list1;
>>
>> >> > I have created a linked list ( list1 )like this 1 -> 2 -> 3 -> 4
>>
>> >> > so i free it like this ----
>>
>> >> > free(list1 -> next -> next ->next);
>> >> > free(list1 -> next -> next);
>> >> > free(list1 -> next);
>> >> > free(list1);
>>
>> >> > when i am printing the list after each free, it is always printing a
>> >> > list of length 4, isn't the values free'd when we do free() ?
>>
>> >> > actual printing gives
>> >> > 1 2 3 0
>> >> > 1 2 garbage 0
>> >> > 1 garbage garbage 0
>> >> > garbage garbage garbage 0
>>
>> >> > why is the linked list still connected ?
>>
>> >> > actual print function -----
>> >> > void print(node *l)
>> >> > {
>> >> > while(l != NULL)
>> >> > {
>> >> > printf("%d\t",l->num);
>> >> > l = l->next;
>> >> > }
>> >> > printf("\n");
>>
>> >> > }
>>
>> >> --
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>> --
>> Saurabh Singh
>> B.Tech (Computer Science)
>> MNNIT ALLAHABAD
>
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