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.
On Fri, Nov 11, 2011 at 8:36 PM, shady <sinv...@gmail.com> wrote:
> ok, thanks.
>
> why do we need to free 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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