bt the ans that sharad gave is ryt..
acc to me 1st row n 1st col of o/p shd b 2 (if size of int is 2) bt it is
1...

On 13 June 2010 12:10, Rohit Saraf <rohit.kumar.sa...@gmail.com> wrote:

> @divya: u r rite.. that * should not be there
>
> --------------------------------------------------
> Rohit Saraf
> Second Year Undergraduate,
> Dept. of Computer Science and Engineering
> IIT Bombay
> http://www.cse.iitb.ac.in/~rohitfeb14<http://www.cse.iitb.ac.in/%7Erohitfeb14>
>
>
> On Sun, Jun 13, 2010 at 11:07 AM, divya jain <sweetdivya....@gmail.com>wrote:
>
>> ptr is a pointer naaa...then why ptr-p=*(&(arr+1)-&arr)  ???
>> why not &(arr+1)-&arr ??
>> i knw m wrong somewhr...plz correct me
>>
>>
>> On 13 June 2010 07:57, Mahesh_JNU <mahesh.jnumc...@gmail.com> wrote:
>>
>>> agreed .........
>>>
>>>
>>> On Sun, Jun 13, 2010 at 7:48 AM, sharad kumar 
>>> <aryansmit3...@gmail.com>wrote:
>>>
>>>> 111
>>>> 222
>>>> 333
>>>> 344
>>>> ptr++ ->u do posst increment
>>>> hence it goes to 1
>>>> ptr-p=*(&(arr+1)-&arr)=1
>>>> llrly for other cases
>>>> when u do *ptr++ due to operator precedence ptr++ is done and then
>>>> dereferenced.
>>>> hence u get 222
>>>> next *++ptr
>>>> the ptr is incremented after dereferencing hence u get 333
>>>> next ++*ptr here the value t ptr s incrementas it is treated as++(*ptr)
>>>> hence u get 3 but others refer to location hence 44
>>>>
>>>>
>>>> On Sat, Jun 12, 2010 at 9:21 PM, divya <sweetdivya....@gmail.com>wrote:
>>>>
>>>>> #include<stdio.h>
>>>>> int main()
>>>>> {
>>>>> static int arr[]={0,1,2,3,4};
>>>>> int *p[]={arr,arr+1,arr+2,arr+3,arr+4};
>>>>> int **ptr=p;
>>>>> ptr++;
>>>>> printf("%d %d %d\n",ptr-p,*ptr-arr,**ptr);
>>>>> *ptr++;
>>>>> printf("%d %d %d\n",ptr-p,*ptr-arr,**ptr);
>>>>> *++ptr;
>>>>> printf("%d %d %d\n",ptr-p,*ptr-arr,**ptr);
>>>>> ++*ptr;
>>>>> printf("%d %d %d\n",ptr-p,*ptr-arr,**ptr);
>>>>> return 0;
>>>>> }
>>>>> wat shd b the o/p n why...
>>>>>
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>>>>
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