如何使用双指针清除链表？

Question

Hi I'm trying to make a function that clears a linked list that *first will point to, then the node **first should be freed and the pointer *first set to NULL .

I'm having trouble grasping double pointers and can't get this to work correctly.

Answer 1

You have to move to the next list element before you delete the node. Otherwise you are accessing memory that has been freed.

while( *first != NULL )
{
    temp = *first;
    *first = temp->next;
    free(temp);
}

Be aware that because you're trashing the whole list, *first is going to eventually be NULL. So you can use a single-pointer (just like temp ) to traverse your list, and then set *first = NULL at the end. That saves an extra pointer indirection, which arguably is wasteful of CPU in this case.

[edit] What I mean is:

struct node *curr = *first;
struct node *prev;            

while( curr != NULL )
{
    prev = curr;
    curr = curr->next;
    free(prev);
}

*first = NULL;

Generally, I find that the less pointer dereferencing you have going on, the easier the code is to understand at a glance.

Answer 2

node* remove_node(node **double_pointer,int search_value)
//pass head of your node as a parameter to double pointer
{while(*double_pointer && **(double_pointer).value!=search_value)
{
double_pointer=&((**double_pointer)->next);
}
if(**double_pointer!=null)
{//lines below are to delete the node which contains our search value
node* deleted_node=*double_pointer;
*double_pointer=*double_pointer->next;
return deleted node;
}}