bi-linked list based on one-way linked list

createbi, creates a bi-linked list out of the one-way linked list.
I don't see how I do these assignments, without failing, when I have 2 types of struct.
Please help.

Code:

---

void createbi (FOO **list, FOO1 **bilist)
{
        FOO *prev, *crnt, *ahd;
        crnt=*list;
       
        while (crnt!=NULL)
                crnt= crnt->next;
        crnt->next=prev;
        *bilist=crnt;
        (*bilist)->num =crnt->num;
        (*bilist)->prev=NULL;
        (*bilist)->next=prev;
        while (bilist!=list)
        {
                ahd=crnt->next;
                (*bilist)->prev=ahd;
                crnt->next=prev;
                (*bilist)->next=prev;
                (*bilist)->num=crnt->num;
                prev=crnt;
                crnt=ahd;
        }

(*bilist)->next=NULL;
}

---

Well first off, lose the second indirection and your code will become a lot cleaner.

Code:

---

void foo( struct sNode **sList, struct dNode **dList )
{
    struct sNode sPtr = *sList;
    struct dNode dPtr = *dList;
    /* Much better already... */
}

---

Wait wait wait wait... Do you already have the second list built? (ie: allocated) If so this is something you can do in your sleep:

Code:

---

void foo( struct sNode **sList, struct dNode **dList )
{
    struct sNode sPtr = *sList;
    struct dNode dPtr = *dList;
    /* Much better already... */

    for( ; sPtr; sPtr = sPtr->next, dPtr = dPtr->next )
    {
        dPtr->data1 = sPtr->data1;
        dPtr->data2 = sPtr->data2;
        ...
      dPtr->dataN = sPtr->dataN;
    }
}

---

That's all folks!

If you don't have it allocated, then you do the exact same thing, except you allocate a node in there for dPtr, and link it to the double linked list you're building.

Otherwise you need to explain what it is you're trying to do.

Quzah.