Re: [algogeeks] address calculation

2011-08-06 Thread Aditya Virmani
1033,2066?

On Sat, Aug 6, 2011 at 11:09 PM, aditi garg aditi.garg.6...@gmail.comwrote:

 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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Re: [algogeeks] address calculation

2011-08-06 Thread aditi garg
No the ans is 1031 and 20C4...
i got it btw...thanx :)

On Sat, Aug 6, 2011 at 11:13 PM, Aditya Virmani virmanisadi...@gmail.comwrote:

 1033,2066?


 On Sat, Aug 6, 2011 at 11:09 PM, aditi garg aditi.garg.6...@gmail.comwrote:

 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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Re: [algogeeks] address calculation

2011-08-06 Thread ankit sambyal
A[3][4]= 1049
B[3][4]= 2196

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Re: [algogeeks] address calculation

2011-08-06 Thread siddharth srivastava
On 6 August 2011 23:15, aditi garg aditi.garg.6...@gmail.com 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 
 virmanisadi...@gmail.comwrote:

 1033,2066?


 On Sat, Aug 6, 2011 at 11:09 PM, aditi garg aditi.garg.6...@gmail.comwrote:

 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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Re: [algogeeks] address calculation

2011-08-06 Thread ankit sambyal
@aditi: explain ur answer.. How u got it ?

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Re: [algogeeks] address calculation

2011-08-06 Thread aditi garg
@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
akssps...@gmail.comwrote:



 On 6 August 2011 23:15, aditi garg aditi.garg.6...@gmail.com 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 virmanisadi...@gmail.com
  wrote:

 1033,2066?


 On Sat, Aug 6, 2011 at 11:09 PM, aditi garg 
 aditi.garg.6...@gmail.comwrote:

 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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Re: [algogeeks] address calculation

2011-08-06 Thread Ram Chauhan
1049 and 1098

On Sat, Aug 6, 2011 at 11:09 PM, aditi garg aditi.garg.6...@gmail.comwrote:

 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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Re: [algogeeks] address calculation

2011-08-06 Thread akshay khatri
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 rb.chau...@gmail.com wrote:

 1049 and 1098

 On Sat, Aug 6, 2011 at 11:09 PM, aditi garg aditi.garg.6...@gmail.comwrote:

 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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Re: [algogeeks] address calculation

2011-08-06 Thread aditi garg
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

On Sat, Aug 6, 2011 at 11:37 PM, akshay khatri akshaykhatri...@gmail.comwrote:

 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 rb.chau...@gmail.com wrote:

 1049 and 1098

 On Sat, Aug 6, 2011 at 11:09 PM, aditi garg aditi.garg.6...@gmail.comwrote:

 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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Re: [algogeeks] address calculation

2011-08-06 Thread akshay khatri
On 6 August 2011 23:40, aditi garg aditi.garg.6...@gmail.com 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 
 akshaykhatri...@gmail.comwrote:

 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 rb.chau...@gmail.com wrote:

 1049 and 1098

 On Sat, Aug 6, 2011 at 11:09 PM, aditi garg 
 aditi.garg.6...@gmail.comwrote:

 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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Re: [algogeeks] address calculation

2011-08-06 Thread aditi garg
@akshay: what i meant ws suppose the array strted from  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  so add the base address and u get
the ans 1031 and 20C4...i hope its clear now...

On Sat, Aug 6, 2011 at 11:49 PM, akshay khatri akshaykhatri...@gmail.comwrote:



 On 6 August 2011 23:40, aditi garg aditi.garg.6...@gmail.com 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 akshaykhatri...@gmail.com
  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 rb.chau...@gmail.com wrote:

 1049 and 1098

 On Sat, Aug 6, 2011 at 11:09 PM, aditi garg 
 aditi.garg.6...@gmail.comwrote:

 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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Re: [algogeeks] address calculation

2011-08-06 Thread akshay khatri
On 6 August 2011 23:56, aditi garg aditi.garg.6...@gmail.com wrote:

 @akshay: what i meant ws suppose the array strted from  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  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 
 akshaykhatri...@gmail.comwrote:



 On 6 August 2011 23:40, aditi garg aditi.garg.6...@gmail.com 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 
 akshaykhatri...@gmail.com 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 rb.chau...@gmail.com wrote:

 1049 and 1098

 On Sat, Aug 6, 2011 at 11:09 PM, aditi garg aditi.garg.6...@gmail.com
  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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Re: [algogeeks] address calculation

2011-08-06 Thread aditi garg
oh @ akshay y ru doing 10x4??? 10 is the no of rows not the columns so it
shud be 15x3+4...and by multiplying by 4(in 10x4) ur taking it to the 5th
row ie A[4][4] its only A[3][4]...
i hope its clear now


On Sun, Aug 7, 2011 at 12:14 AM, akshay khatri akshaykhatri...@gmail.comwrote:



 On 6 August 2011 23:56, aditi garg aditi.garg.6...@gmail.com wrote:

 @akshay: what i meant ws suppose the array strted from  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  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 akshaykhatri...@gmail.com
  wrote:



 On 6 August 2011 23:40, aditi garg aditi.garg.6...@gmail.com 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 
 akshaykhatri...@gmail.com 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 rb.chau...@gmail.com wrote:

 1049 and 1098

 On Sat, Aug 6, 2011 at 11:09 PM, aditi garg 
 aditi.garg.6...@gmail.com 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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Aditi Garg
Undergraduate Student
Electronics  Communication Divison
NETAJI SUBHAS INSTITUTE OF TECHNOLOGY
Sector 3, Dwarka
New Delhi

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Re: [algogeeks] address calculation

2011-08-06 Thread rajeev bharshetty
@aditi : I Think the output depends on whether it is row major or column
major representation in memory .
 Does he row amjor or column

On Sun, Aug 7, 2011 at 12:19 AM, aditi garg aditi.garg.6...@gmail.comwrote:

 oh @ akshay y ru doing 10x4??? 10 is the no of rows not the columns so it
 shud be 15x3+4...and by multiplying by 4(in 10x4) ur taking it to the 5th
 row ie A[4][4] its only A[3][4]...
 i hope its clear now



 On Sun, Aug 7, 2011 at 12:14 AM, akshay khatri 
 akshaykhatri...@gmail.comwrote:



 On 6 August 2011 23:56, aditi garg aditi.garg.6...@gmail.com wrote:

 @akshay: what i meant ws suppose the array strted from  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  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 
 akshaykhatri...@gmail.com wrote:



 On 6 August 2011 23:40, aditi garg aditi.garg.6...@gmail.com 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 
 akshaykhatri...@gmail.com 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 rb.chau...@gmail.com wrote:

 1049 and 1098

 On Sat, Aug 6, 2011 at 11:09 PM, aditi garg 
 aditi.garg.6...@gmail.com 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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 --
 Aditi Garg
 Undergraduate Student
 Electronics  Communication Divison
 NETAJI SUBHAS INSTITUTE OF TECHNOLOGY
 Sector 3, Dwarka
 New Delhi


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