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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>
>
> --
> Mahesh Giri
> MCA Final Sem
> JNU, New Delhi
>
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