array pointer

array pointer

[ratheesh k]

int  *s_ptr;
int  *d_ptr;
int arr[2][2]={1,2,3,4};

s_ptr=*arr;  /* This is perfectly fine */
d_ptr=arr /* this line throws a  warning: assignment from incompatible
pointer type */

What is the problem here ?

Re: array pointer

[ratheesh k]

i made a mistake  d_ptr is  int **d_ptr

On Tue, Jan 18, 2011 at 6:52 PM, ratheesh k <ratheesh.ksz@gmail.com> wrote:
> int  *s_ptr;
> int  *d_ptr;
> int arr[2][2]={1,2,3,4};
>
> s_ptr=*arr;  /* This is perfectly fine */
> d_ptr=arr /* this line throws a  warning: assignment from incompatible
> pointer type */
>
> What is the problem here ?
>
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Re: array pointer

[ratheesh k]

int  *s_ptr;
int  **d_ptr;
int arr[2][2]={1,2,3,4};

s_ptr=*arr;  /* This is perfectly fine */
d_ptr=arr /* this line throws a  warning: assignment from incompatible
pointer type */  ????????



On Tue, Jan 18, 2011 at 6:53 PM, ratheesh k <ratheesh.ksz@gmail.com> wrote:
> i made a mistake  d_ptr is  int **d_ptr
>
> On Tue, Jan 18, 2011 at 6:52 PM, ratheesh k <ratheesh.ksz@gmail.com> wrote:
>> int  *s_ptr;
>> int  *d_ptr;
>> int arr[2][2]={1,2,3,4};
>>
>> s_ptr=*arr;  /* This is perfectly fine */
>> d_ptr=arr /* this line throws a  warning: assignment from incompatible
>> pointer type */
>>
>> What is the problem here ?
>>
>
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Re: array pointer

[Michal Nazarewicz]

On Tue, 18 Jan 2011 14:32:43 +0100, ratheesh k <ratheesh.ksz@gmail.com>  
wrote:

> int  *s_ptr;
> int  **d_ptr;
> int arr[2][2]={1,2,3,4};

One image is worth a thousand words, so here it goes:

arr is something like this in memory:

+---+---+---+---+
| 1 | 2 | 3 | 4 |
+---+---+---+---+

d_ptr would require something like:

+---------+---------+
| pointer | pointer |
+----|----+----|----+
      |         |     +---+---+
      |         +---> | 3 | 4 |
      |     +---+---+ +---+---+
      +---> | 1 | 2 |
            +---+---+

Ie. in arr all the numbers are lied linearly in memory whereas
int **d_ptr is an array of pointers to array of ints.

Re: array pointer

[David McMurray]

>> int  *s_ptr;
>> int  **d_ptr;
>> int arr[2][2]={1,2,3,4};
>
> One image is worth a thousand words, so here it goes:
>
> arr is something like this in memory:
>
> +---+---+---+---+
> | 1 | 2 | 3 | 4 |
> +---+---+---+---+
>
> d_ptr would require something like:
>
> +---------+---------+
> | pointer | pointer |
> +----|----+----|----+
>     |         |     +---+---+
>     |         +---> | 3 | 4 |
>     |     +---+---+ +---+---+
>     +---> | 1 | 2 |
>           +---+---+
>
> Ie. in arr all the numbers are lied linearly in memory whereas
> int **d_ptr is an array of pointers to array of ints.

I would say int **d_ptr is really a pointer to a pointer and not
necessarily an array of pointers.

So, the following is fine:
d_ptr = &s_ptr;
printf("s_ptr: %d, d_ptr: %d\n", *s_ptr, **d_ptr);

and displays "s_ptr: 1, d_ptr: 1"

To (attempt to) answer your original question though, the problem is
that "arr" will give you the address of the array, not the address of
a pointer to the array. For each extra dimension of the array you need
to add a * when referring to the name of the array without indices in
order to de-reference the additional dimensions.

So,
1-dimensional array
int *s_ptr;
int arr[2]={1,2};
s_ptr = arr; /* Ok because arr references the address of the array
immediately */

2-dimensional array
int *s_ptr;
in arr[2][2]={1,2,3,4};
s_ptr = *arr; /* Ok because the * de-references arr to the address of
the array, but arr is not the address of a pointer to the array */

3-dimensional array
int *s_ptr;
int arr[2][2][2]={1,2,3,4,5,6,7,8};
s_ptr = **arr; /* Ok, again each * de-references additional dimensions
of the array */

But these references are not pointers, which is why int **d_ptr; int
arr[2][2]={1,2,3,4}; d_ptr = arr; will not work, arr will not be a
pointer to a pointer.

HTH,
David McMurray.
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Re: array pointer

[Glynn Clements]

ratheesh k wrote:

> int  *s_ptr;
> int  **d_ptr;
> int arr[2][2]={1,2,3,4};
> 
> s_ptr=*arr;  /* This is perfectly fine */
> d_ptr=arr /* this line throws a  warning: assignment from incompatible
> pointer type */  ????????

Arrays are not pointers. If you use the name of an array as an
expression, it is treated as a pointer to the first element.

For one-dimensional arrays, this means that you can use an array in
most places where a pointer is required. But this doesn't hold for
higher dimensions; an array of pointers isn't the same as an array of
arrays. It also doesn't hold for many other cases, i.e. sizeof,
variable declarations, etc.

-- 
Glynn Clements <glynn@gclements.plus.com>