Re: declaring array of pointers
From: Al Bowers (xabowers_at_rapidsys.com)
Date: 09/19/04
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Date: Sun, 19 Sep 2004 09:12:21 -0400
Steve Lambert wrote:
> Hi,
>
> I'd be grateful if someone could clarify this for me. In the linked list
> structure my intention is to declare an array of length 3 containing
> pointers to node
> eg. Node *Iterators[3]
> The compiler seems to interpret this as a pointer to an array of 3 nodes
> instead. This interpretation ensures that the second assigment to mynode
> below fails compilation with the given message.
>
> Could someone explain to me how to correctly declare an array of length 3
> containing pointers to node
>
> cheers
>
>
> #include <stdio.h>
>
> typedef struct Node
> {
> void *Data;
> struct Node *Next;
> struct Node *Previous;
> } Node;
>
> /* structure to represent a linked list */
> typedef struct LinkedList
> {
> Node *Head; /* start of linked list */
> Node *Tail; /* end of linked list */
> long NumberOfNodes;
> short NumberOfIterators;
> Node *Iterators[3];
> } LinkedList;
>
> LinkedList *mLinkedLists[2];
>
>
> void main()
> {
> Node *mynode;
> LinkedList *LL=NULL;
>
> LL = mLinkedLists[0];
>
> mynode = LL->Iterators[0]; <- this compiles
>
It may compile but it is very flawed and will not run.
LL has the value NULL and does not point to storage.
Since you are trying to access storage that is not, the
execution will fail.
You can type the member Iterators as
Node *Iterators[3];
making Iterators point to an array of 3 Node pointers.
Make a LinkedList object
LinkledList mylist;
Then you can access or assign Iterators with code something
like this:
mylist.Iterators[0] = malloc(sizeof Node);
or
Node newnode;
mylist.Iterators[0] = &newnode;
and
mylist.Iterators[0]->Data = /* whatever */
One the other hand out might make the member Iterators:
Node (*Iterators)[3];
Iterators will point to an array of 3 Node objects (not pointers).
Here is an example, in function main below:
#include <stdio.h>
#include <stdlib.h>
typedef struct Node
{
int Data;
struct Node *Next;
struct Node *Previous;
} Node;
/* structure to represent a linked list */
typedef struct LinkedList
{
Node *Head; /* start of linked list */
Node *Tail; /* end of linked list */
size_t NumberOfNodes;
size_t NumberOfIterators;
Node (*Iterators)[3];
} LinkedList;
/* Prototypes */
int AddNode(LinkedList *p, int data, size_t IterNo);
void FreeLinkedList(LinkedList *p);
int main(void)
{
LinkedList myList = {NULL};
size_t i,j;
AddNode(&myList, 44,0);
AddNode(&myList,33,2);
printf("myList.NumberOfIterators = %u\n",
myList.NumberOfIterators);
for(i = 0; i < myList.NumberOfIterators;i++)
for(j = 0; j < 3; j++)
{
myList.Iterators[i][j].Data = i+j;
printf("myList.Iterator[%u][%u].Data = %d\n",
i,j, myList.Iterators[i][j].Data);
}
printf("\nmyList.Head->Data = %d\n",myList.Head->Data);
FreeLinkedList(&myList);
return 0;
}
int AddNode(LinkedList *p, int data, size_t IterNo)
{
Node *new;
Node (*itmp)[3];
if((new = malloc(sizeof *new)) == NULL) return 0;
itmp = realloc(p->Iterators,(sizeof *itmp)*IterNo);
if(IterNo && itmp == NULL)
{
free(new);
return 0;
}
p->Iterators = itmp;
p->NumberOfIterators = IterNo;
new->Data = data;
if(p->NumberOfNodes == 0)
{
new->Previous = new->Next = NULL;
p->Head = p->Tail = new;
}
else
{
new->Previous = NULL;
new->Next = p->Head;
p->Head->Previous = new;
p->Head = new;
}
p->NumberOfNodes++;
return 1;
}
void FreeLinkedList(LinkedList *p)
{
Node *tmp;
for(tmp = p->Head; tmp;p->Head = tmp)
{
tmp = p->Head->Next;
free(p->Head);
}
free(p->Iterators);
p->Head = p->Tail = NULL;
p->Iterators = NULL;
p->NumberOfIterators = p->NumberOfNodes = 0;
return;
}
-- Al Bowers Tampa, Fl USA mailto: xabowers@myrapidsys.com (remove the x to send email) http://www.geocities.com/abowers822/
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