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

ans is  1344

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

it should be 1330 according to me

Hi ,

Can any one explain this with proper steps

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.

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.

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

The answer is

1) If it is a row based then

1258+2(9(5-1)+(8-1))=1258+2(36+7)=1258+86=1344

2)
If it is column based then

1258+2((7(8-1)+(5-1))=1258+2(49+4)=1258+106=1364.

1258+4*(9*(5-1)+(8-1)) equals to 1430