On 6 August 2011 23:56, aditi garg <[email protected]> wrote:

> @akshay: what i meant ws suppose the array strted from 0000 in decimal...so
> A[3][4] wud be at 0049 and B[3][4] will be at 0196
> convert dese 2 values to hex it gives 31 and C4 respectively...now base
> address is 1000 and 2000 instead if 0000 so add the base address and u get
> the ans 1031 and 20C4...i hope its clear now...
>

I still don't get how you got the figures of 49 and 196

I calculated as follows
A[3][4] = 10x4 + 4 = 44 bytes
B[3][4] = (10x4 + 4 )x4 = 176

>
>
> On Sat, Aug 6, 2011 at 11:49 PM, akshay khatri 
> <[email protected]>wrote:
>
>>
>>
>> On 6 August 2011 23:40, aditi garg <[email protected]> wrote:
>>
>>> A[3][4] wud be in the 4th row...so strtung address of 4th row wud be
>>> 46..and thn 4th element wud be at 49...similarly fr B 180 fr the frst 3 rows
>>> + 16 fr the 4th elemnet so 196
>>>
>>
>> How does it start from 46 and 180 ?
>> as per my knowledge, 0x1000 = 4096 and 0x2000 is 8192
>>
>>
>>>
>>>  On Sat, Aug 6, 2011 at 11:37 PM, akshay khatri <
>>> [email protected]> wrote:
>>>
>>>> how is that 49 bytes and 196 bytes
>>>> shouldn't it be 44 and 176 bytes respectively
>>>>
>>>>
>>>> On 6 August 2011 23:26, Ram Chauhan <[email protected]> wrote:
>>>>
>>>>> 1049 and 1098
>>>>>
>>>>> 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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>
>
>
> --
> Aditi Garg
> Undergraduate Student
> Electronics & Communication Divison
> NETAJI SUBHAS INSTITUTE OF TECHNOLOGY
> Sector 3, Dwarka
> New Delhi
>
>
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