thnks all :)
@rajeev : u are right dats y  *((char*)iPtr+2) will printf 0.

On Mon, Jul 25, 2011 at 11:56 AM, ~*~VICKY~*~ <[email protected]>wrote:

> Consider the binary representation of 257 which is 100000001
>
> which will be stored in little endain representation as least significant
> eight bits + most significant eight bits as follows
>
> 00000001 | 00000001
>
> now iptr on casting to char will point to least significant eight bits
> which is 1, when u increment iptr it refers to most sig 8 bits which is also
> 1
>
> hence the o/p will be : 1 1
>
>
> hope it helps!
>
> On Mon, Jul 25, 2011 at 11:41 AM, aditya kumar <
> [email protected]> wrote:
>
>> main()
>> {
>> int i = 257;
>> int *iPtr = &i;
>> printf("%d %d", *((char*)iPtr), *((char*)iPtr+1) );
>> }
>>
>> can any one explain me the o/p ??
>>
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>
>
> --
> Cheers,
>
>   Vicky
>
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