i think 1st is wrong bt 2nd is correct......

On Fri, Aug 5, 2011 at 9:51 PM, nishaanth <[email protected]> wrote:

> 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 <[email protected]>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 <
>> [email protected]> 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 
>>> <[email protected]>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 <
>>>> [email protected]> 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?
>>>>> --
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>>>
>>>
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>
>
>
> --
> S.Nishaanth,
> Computer Science and engineering,
> IIT Madras.
>
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-- 

      Regards
Himanshu Kansal
  Msc Comp. sc.
(University of Delhi)

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