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?
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