@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
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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>>>
>>>
>>> --
>>> yezhu malai vaasa venkataramana Govinda Govinda
>>>
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>>
>>
>>
>> --
>> Mahesh Giri
>> MCA Final Sem
>> JNU, New Delhi
>>
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