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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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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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]= 1049
B[3][4]= 2196
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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: explain ur answer.. How u got it ?
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@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]
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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You received this message because you are subscribed to the Google
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
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
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
@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
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
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
@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
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