depends on endianess of processor

On Mon, Jul 25, 2011 at 12:22 PM, aditya kumar <[email protected]
> wrote:

> 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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