i think both are erroneous.
first statement .... i think you are trying to change the array address
which is not possible.
second statement....&arr doesn't make any sense i guess.... arr gives the
address but &arr is not allowed
On Fri, Aug 5, 2011 at 4:19 PM, Vijay Khandar <vijaykhand...@gmail.com>wrote:
> but u initialized **dp means .....
> ex-dp=&p and p=&arr then its correct so dp contains addr of p which
> inturns contains addrof arr now **dp is correct initialization.
>
>
> On Fri, Aug 5, 2011 at 7:45 PM, Arun Vishwanathan
> <aaron.nar...@gmail.com>wrote:
>
>> i see but is not arr a pointer to first array element and so &arr contain
>> address of that pointer ?
>>
>>
>> On Fri, Aug 5, 2011 at 4:06 PM, Vijay Khandar <vijaykhand...@gmail.com>wrote:
>>
>>> I dont think so dp=&arr; since **dp; dp contains the addr of another
>>> ptr variable...
>>>
>>> On Fri, Aug 5, 2011 at 7:27 PM, Arun Vishwanathan <
>>> aaron.nar...@gmail.com> wrote:
>>>
>>>>
>>>> I guess someone had posted a link earlier from which I have a basic
>>>> doubt
>>>>
>>>> when u have
>>>>
>>>> int arr[3]={1,0,2};
>>>> int **dp;
>>>> int (*pa)[3];
>>>>
>>>> is this the right assingment for instance?
>>>>
>>>> pa=arr;
>>>> dp=&arr;
>>>>
>>>> or have I flipped the ampersand in assigning?
>>>>
>>>> Also when I do pa++ will it jump by size of int or the whole array
>>>> size it points to?
>>>> --
>>>> "People often say that motivation doesn't last. Well, neither does
>>>> bathing - that's why we recommend it daily."
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>>
>>
>>
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--
S.Nishaanth,
Computer Science and engineering,
IIT Madras.
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