Aptitude - 4.In two dimensional array X(7,9) each element occupies 2 bytes of memory.If the address of first el

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4.In two dimensional array X(7,9) each element occupies 2 bytes of memory.If the address of first element X(1,1)is 1258 then what will be the address of the element X(5,8) ?
whats the ans

Total Answers and Comments: 7

Last Update: April 26, 2008 Asked by: pavan

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September 28, 2006 04:08:35

vinayak

RE: 4.In two dimensional array X(7,9) each element occ...

ans is 1344

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September 30, 2006 08:50:08

naveen

RE: 4.In two dimensional array X(7,9) each element occ...

Ans is 1344 and it is calculated asfirst add 8 since in first row 8 elements then for the next 3 row add 9 nd for the last row add 8 so the total comes to 43 multiply with 2 add it to 1258 it gives 1344

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October 05, 2006 00:39:07

ajay

RE: 4.In two dimensional array X(7,9) each element occ...

it should be 1330 according to me

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October 05, 2006 21:09:21

joseprabhu

Aspirant.

Hi ,

Can any one explain this with proper steps

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November 30, 2006 11:03:20

vishal

RE: 4.In two dimensional array X(7,9) each element occ...

intger req. 2 byte for memory allocation so in this two dim array 7x9 it req. 126 bytes memory.

so 1x1 starting add is 1258.

now to find add of 5x8 so it req 80 byes

so 80+1258=1338.

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January 24, 2007 04:30:13

prashant kara

RE: 4.In two dimensional array X(7,9) each element occ...

here, the address of first element x[1][1] is 1258 and also 2 byte of memory is given.now, we have to solve the address of element x[5][8], therefore, 1258+ 5*8*2 = 1258+80 = 1338 so the answer is 1338.

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April 26, 2008 22:19:55

imran4108

Member Since: April 2008 Contribution: 1

RE: 4.In two dimensional array X(7,9) each element occupies 2 bytes of memory.If the address of first element X(1,1)is 1258 then what will be the address of the element X(5,8) ? whats the ans

formulae is
X(5,8)=X(N,M)
X(7,9)=X(R,C)
X(5,8)=BASE+C*(M-1)*SIZE+(N-1)*SIZE
WHERE SIZE=2BYTES
BASE=1258
X(5,8)=1258+9*7*2+4*2

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