Two-Dimensional Arrays

A Two-dimensional array X (7,9) is stored linearly column-wise in a computer's 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)?

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Sandhya.Kishan  

  • Member Since Mar-2012 | Apr 17th, 2012


use 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

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