Page 310 - Computer science 868 Class 12
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= 1500 + 1 * [160]
                       = 1660
                    (ii)  Row Major Wise Calculation of the above equation
                       The given values are: B = 1500, W = 1 byte, I = 15, J = 20, Lr = -15, Lc = 15, N = 26
                       Address of A [ I ][ J ] = B + W * [ N * ( I - Lr ) + ( J - Lc ) ]

                       = 1500 + 1* [26 * (15 - (-15))) + (20 - 15)]
                       = 1500 + 1 * [26 * 30 + 5]
                       = 1500 + 1 * [780 + 5]
                       = 1500 + 785
                       = 2285
                2.   Given an array, arr[1………10][1………15] with base value 100 and the size of each element is 1 Byte in memory.
                    Find the address of arr[8][6] with the help of row-major order.
               Solution:
                     Given:
                    B   =  100
                    W  =  1 Bytes
                    I   =  8
                    J   =  6
                    LR  =  1
                    LC  =  1
                    Number of columns given in the matrix N = Upper Bound – Lower Bound + 1
                       = 15 - 1 + 1
                       = 15
                    Formula:
                    Address of A[I][J] = B + W * ((I - LR) * N + (J - LC))
               Ans.  Address of A[8][6]

                    = 100 + 1 * ((8 - 1) * 15 + (6 - 1))
                    = 100 + 1 * ((7) * 15 + (5))
                    = 100 + 1 * (110)
                    = 210

              Questions related to Address Calculations
                1.   Given the base address of an array B[1400…..1800] as 1020 and the size of each element is 2 bytes in the
                    memory. Find the address of B[1700].
                2.   Given the base address of array B[1500…..3000] as 1260 and the size of each element is 2 bytes in the memory.
                    Find the address of B[2200] and B[2000].
                3.   The array D[-2…10][3…8] contains double type elements. If the base address is 4110. Find the address D[4][5],
                    when the array is stored in column major wise.
                4.   An array AR[-4…6, -2 … 12] stores elements in Row Major wise, with the address of AR[2][3] as 4142. If each
                    element requires 2 bytes of storage. Find the base address.
                5.   A matrix B[10][7] is stored in the memory with each element requiring 2 bytes of storage. If the base address at B[x]
                    [1] is 1012 and the address at B[7][3] is 1060. Determine the value of ‘x’ where the matrix is stored column major
                    wise.




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