A Two-dimensional array X (7,9) is stored linearly column-wise in a computers memory. Each element requires 8 bytes for storage of the value. If the first byte address of X (1,1) is 3000, what would be the last byte address of X (2,3)?
Submitted by: Administratoruse the formulae
X(i,j)=Base+w[n(i-1)+(j-1)]
where m=7 ,n =9 ,i=2 ,j=3
hence 3000+8*[9(2-1)+(3-1)]
=3000+8*(9+2)
=3000+8*11=3088
Submitted by: Administrator
X(i,j)=Base+w[n(i-1)+(j-1)]
where m=7 ,n =9 ,i=2 ,j=3
hence 3000+8*[9(2-1)+(3-1)]
=3000+8*(9+2)
=3000+8*11=3088
Submitted by: Administrator
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