@ankit ur ans is fine bt u missed a point that dey are calculating in hex so
convert ur ans to hex and it will be the same as the one i posted...
@siddharth...A[3][4] will be 49 bytes ahead of the base address...and 49 in
hex is 31 so ans will be 1031
and int (assuming 4 bytes) the value B[3][4] will be 196 bytes ahead of base
address whch in hex is C 4 hence the ans 20C4...
I hope its clear

On Sat, Aug 6, 2011 at 11:18 PM, siddharth srivastava
<[email protected]>wrote:

>
>
> On 6 August 2011 23:15, aditi garg <[email protected]> wrote:
>
>> No the ans is 1031 and 20C4...
>> i got it btw...thanx :)
>>
>
> explain ?
>
>>
>>
>> On Sat, Aug 6, 2011 at 11:13 PM, Aditya Virmani <[email protected]
>> > wrote:
>>
>>> 1033,2066?
>>>
>>>
>>> On Sat, Aug 6, 2011 at 11:09 PM, aditi garg 
>>> <[email protected]>wrote:
>>>
>>>> CHAR A[10][15] AND INT B[10][15] IS DEFINED
>>>> WHAT'S THE ADDRESS OF A[3][4] AND B[3][4]
>>>> IF ADDRESS OF A IS OX1000 AND B IS 0X2000
>>>>
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>>
>>
>> --
>> Aditi Garg
>> Undergraduate Student
>> Electronics & Communication Divison
>> NETAJI SUBHAS INSTITUTE OF TECHNOLOGY
>> Sector 3, Dwarka
>> New Delhi
>>
>>
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> --
> Regards
> Siddharth Srivastava
>
>
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-- 
Aditi Garg
Undergraduate Student
Electronics & Communication Divison
NETAJI SUBHAS INSTITUTE OF TECHNOLOGY
Sector 3, Dwarka
New Delhi

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