Before looking at the answer, try writing a simple C program (with a for loop) to do this. Quite a few people get this wrong.
This is the wrong way to do it
struct list *listptr, *nextptr;
for(listptr = head; listptr != NULL; listptr = listptr->next)
{
free(listptr);
}
If you are thinking why the above piece of code is wrong, note that once you free the listptr node, you cannot do something like listptr = listptr->next!. Since listptr is already freed, using it to get listptr->next is illegal and can cause unpredictable results!
This is the right way to do it
struct list *listptr, *nextptr;
for(listptr = head; listptr != NULL; listptr = nextptr)
{
nextptr = listptr->next;
free(listptr);
}
head = NULL;
After doing this, make sure you also set the head pointer to NULL!
Wrong implementation because just a freed pointer can not be a NULL at any point of time but an invlid pointer.
ReplyDeleteRight one is:
struct list *listptr, *nextptr;
listptr = head;
while (listptr != NULL)
{
nextptr = listptr;
listptr = listptr->next;
free(nextptr);
}
nextptr = NULL;
Wrong implementation because just a freed pointer can not be a NULL at any point of time but an invlid pointer.
ReplyDeleteRight one is:
struct list *listptr, *nextptr;
listptr = head;
while (listptr != NULL)
{
nextptr = listptr;
listptr = listptr->next;
free(nextptr);
}
nextptr = NULL;
Basavaraj : The above code is also correct because:
ReplyDeletehere in for statement
for(listptr = head; listptr != NULL; listptr = nextptr)
it is again assiging a value as listptr = nextptr.
note:free(listptr) will free the adress containing by listptr not the pointer listptr
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