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declaring array of pointers

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

mynode = LL->Iterators; <- this doesn't : cannot convert from 'struct Node
*[3]' to 'struct Node *

printf("out\n");
}
Nov 14 '05 #1
8 2371
"Steve Lambert" <st***********@ntlworld.com> writes:
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]
That's the correct syntax.
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.
It's a misunderstanding of the error message, not a problem with
the compiler.
/* structure to represent a linked list */
typedef struct LinkedList
{ .... Node *Iterators[3];
} LinkedList;
....
Node *mynode;
LinkedList *LL=NULL;

LL = mLinkedLists[0];

mynode = LL->Iterators[0]; <- this compiles
mynode is type Node *.
LL->Iterators[0] is type Node *.
Fine.
mynode = LL->Iterators; <- this doesn't : cannot convert from 'struct Node
*[3]' to 'struct Node *


mynode is type Node *.
LL->Iterators is type Node *[3], just as you declared it.
Problem: you can't assign a Node *[3] to a Node *.

The compiler's error message is slightly deceptive, because the
name of an array object is usually converted into a pointer to
the first element of the array. In this case, that means that
the actual type of the value being assigned is Node **. If you
defined mynode to be of type Node **, the second assignment would
be fine (although the first assignment would become invalid).

What are you trying to do with that statement?
--
"Give me a couple of years and a large research grant,
and I'll give you a receipt." --Richard Heathfield
Nov 14 '05 #2
Steve Lambert wrote:
Node *mynode;
mynode = LL->Iterators[0]; <- this compiles
mynode = LL->Iterators; <- this doesn't : cannot convert from 'struct Node
*[3]' to 'struct Node *


Your essential problem here is that "mynode" is a Node*, and
LL->Iterators is an array of Node*. These types are naturally
incompatible, just as you can't assign an array of int into an int (an
array is not an integer.)

On the other hand, you *can* assign an array of int to an int*. This is
because when used in expression contexts, the name of an array "decays"
into a pointer to its first element implicitly. In your case,
LL->Iterators would "decay" into &LL->Iterators[0], which has type
Node** (which is still incompatible with Node* mynode).
--
Derrick Coetzee
I grant this newsgroup posting into the public domain. I disclaim all
express or implied warranty and all liability. I am not a professional.
Nov 14 '05 #3
"Steve Lambert" <st***********@ntlworld.com> wrote in message news:<FM***************@newsfe5-win.ntli.net>...
... In the linked list
structure my intention is to declare an array of length 3 containing
pointers to node
eg. Node *Iterators[3]
You did. "Size 3" is more idiomatic than "length 3" I think.

<snip> mynode = LL->Iterators[0]; <- this compiles "//" and "/* */" are syntax for comments
-- use them even for meta-comments!
mynode = LL->Iterators; <- this doesn't


Whatever and Whatever[0] are always of different types
(the former often a pointer to the latter) -- why would
you expect *both* statements to compile?

James
Nov 14 '05 #4


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: xa******@myrapidsys.com (remove the x to send email)
http://www.geocities.com/abowers822/

Nov 14 '05 #5
James Dow Allen wrote:

<snip>
mynode = LL->Iterators[0]; <- this compiles

"//" and "/* */" are syntax for comments
-- use them even for meta-comments!


Don't use // in newsgroups, nor for code expected to compiler
under C90.

--
Chuck F (cb********@yahoo.com) (cb********@worldnet.att.net)
Available for consulting/temporary embedded and systems.
<http://cbfalconer.home.att.net> USE worldnet address!
Nov 14 '05 #6
James Dow Allen wrote:

"Steve Lambert" <st***********@ntlworld.com> wrote in message news:<FM***************@newsfe5-win.ntli.net>...
... In the linked list
structure my intention is to declare an array of length 3 containing
pointers to node
eg. Node *Iterators[3]


You did. "Size 3" is more idiomatic than "length 3" I think.


I agree.
Size is a dimension of objects.
Length is a dimension of strings.

--
pete
Nov 14 '05 #7
Derrick Coetzee <dc****@moonflare.com> writes:
On the other hand, you *can* assign an array of int to an int*. This is
because when used in expression contexts, the name of an array "decays"
into a pointer to its first element implicitly. In your case,
LL->Iterators would "decay" into &LL->Iterators[0], which has type
Node** (which is still incompatible with Node* mynode).


I'm curious to know if anyone can shed some light as to the origin of
the phrase "decays into a pointer" when talking about the behavior
of array names used in C. It isn't in K&R (at least not that I
recall), and I don't remember seeing it in any of the other C books
I've read. Personally I find the term misleading, and suspect it
probably confuses C novices more than it helps them. Anyone have
any personal experiences, either pro or con, to report on that?

Nov 14 '05 #8


Tim Rentsch wrote:
I'm curious to know if anyone can shed some light as to the origin of
the phrase "decays into a pointer" when talking about the behavior
of array names used in C.

?
My first experience with the phrase is in the comp.lang.c faq.
Question 6.3 located at
http://www.eskimo.com/~scs/C-faq/q6.3.html

the faq guestion cites the three exceptions.

--
Al Bowers
Tampa, Fl USA
mailto: xa******@myrapidsys.com (remove the x to send email)
http://www.geocities.com/abowers822/

Nov 14 '05 #9

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