Using malloc to create a linked list - malloc

Question

i used malloc in order to allocate new nodes in the list,yet i am facing an error with a certain part of my code;
the following solution applies only to deleting and inserting
#include <stdio.h>
#include <malloc.h>
struct Node{
int value;
struct Node * Next;
struct Node * Previous;
};
typedef struct Node Node;
struct List{
int Count;
int Total;
Node * First;
Node * Last;
};
typedef struct List List;
List Create();
void Add(List a,int value);
void Remove(List a,Node * b);
List Create()
{
List a;
a.Count=0;
return a;
}
void Add(List a,int value)
{
Node * b = (Node *)malloc(sizeof(Node));
if(b==NULL)
printf("Memory allocation error \n");
b->value=value;
if(a.Count==0)
{
b->Next=NULL;
b->Previous=NULL;
a.First=b;
}
else
{
b->Next=NULL;
b->Previous=a.Last;
a.Last->Next=b;
}
++a.Count;
a.Total+=value;
a.Last=b;
}
void Remove(List a,Node * b)
{
if(a.Count>1)
{
if(a.Last==b)
{
b->Previous->Next=NULL;
}
else
{
b->Previous->Next=b->Next;
b->Next->Previous=b->Previous;
}
}
free(b);
}
in the delete function,in the last else condition,i am not certain whether or not using b->Next->Previous is okay,and will work;when using the -> operator,am i adressing to the node pointer or to it's value?

Answer 1

The short answer: Yes, b->Next->Previous is fine -- it's a struct Node*, just like the right hand side b->Previous.
I think that your error lies with the handling of Count: It is incremented by Add(), but Remove() doesn't decrement it. In fact, as the list itself only needs to know whether it is empty or not, you can remove it and instead see if a.First == NULL. (Your a.Count == 1 test can likewise be replaced with a.First != NULL && a.First->Next == NULL test.)
If you're promising Count in you API, you can add it back when you've got the list itself working. The same "remove-then-add-back" might be useful with Total. Think of both of these as caches.
An even better solution would be to implement a circular list:
struct List
{
Node Anchor;
//...
};
List Create()
{
List l;
l.Anchor.Next = l.Anchor.Previous = &l;
return l;
}
bool IsEmpty(List const* l)
{
// Both or neither point at 'l'.
assert((l->Anchor.Next == l) == (l->Anchor.Previous == l));
return l->Anchor.Next == l;
}
// Add a node 'n' to some list after 'ln'.
void AddAfter(Node* n, Node* ln)
{
n->Previous = ln;
n->Next = ln->Next;
n->Next->Previous = n->Previous->Next = n;
}
Node* Remove(Node* n)
{
n->Previous->Next = n->Next;
n->Next->Previous = n->Previous;
n->Next = n->Previous = n; // nice and proper
return x;
}
Now you longer need special cases for empty lists. I let Remove() return the node itself, to make it easy to either move nodes between lists (AddAfter(Remove(somenode), &otherlist.Anchor)) or remove and delete notes (free(Remove(somenode))).
One wart here is that my Anchor node now wastes space for data that never will be used -- but that is easily fixable.