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An n dimensional matrix can be of any dimension. Adding a dimension is adding one more index number (to access the element). In 1-D array you the elements are linearly arranged and can be addressed as a, a, .. . in 2-D array elements are logically in the form of matrix having row and column index and... read more

An n dimensional matrix can be of any dimension. Adding a dimension is adding one more index number (to access the element). In 1-D array you the elements are linearly arranged and can be addressed as a[0], a[1], .. . in 2-D array elements are logically in the form of matrix having row and column index and can be represented as a[0][0], a[0][1] etc. and they are stored row wise in the memory, because the memory is linear. Now a 3-D array is nothing but a logical structure which can be accessed by 3 index numbers. If the array is of size 3 in all the dimensions(int a[3][3][3] then it is stored in the memory in the following order a[0][0][0], a[0][0][1], a[0][0][2], a[0][1][0], a[0][1][1], a[0][1][2], a[0][2][0], a[0][2][1], a[0][2][2], a[1][0][0], a[1][0][1] and so on. Now to find the address of any element of the array based on the given index number and given dimension size, given element size(data type size) and given base address: Suppose the array is of n-dimension having the size of each dimension as S1,S2,S3. . .. Sn And the element size is ES, Base Address is BA And the index number of the element to find the address is given as i1, i2, i3, . . . .in Then the formula will be: Address of A[i1][ i2][ i3]. . . .[ in] = BA + ES*[ i1*( S2* S3* S4 *. . ..* Sn) + i2*( S3* S4* S5 *.. .. * Sn) + .. . . .. + in-2*( Sn-1*Sn) + in-1*Sn + in ] Example-1: An int array A[100][50] is stored at the address 1000. Find the address of A[40][25]. Solution: BA=1000, ES=2, S1=100, S2=50, i1=40, i2=25 Address of A[40][25]=1000+2*[40*50 + 25]=1450 Example-2: An int array A[50][40][30] is stored at the address 1000. Find the address of A[40][20][10]. Solution: BA=1000, ES=2, S1=50, S2=40, S3=30, i1=40, i2=20, i3=10 Address of A[40][20][10]=1000+2*[40*40*30 + 20*30 + 10]=98220 read less

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Array of an element of an array say A is calculated using the following formula: Address of A = B + W * ( I -- LB ) Where, B = Base address W = Storage Size of one element stored in the array (in byte) I = Subscript of element whose address is to be found LB = Lower limit / Lower Bound of... read more

Array of an element of an array say “A[ I ]” is calculated using the following formula: Address of A [ I ] = B + W * ( I – LB ) Where, B = Base address W = Storage Size of one element stored in the array (in byte) I = Subscript of element whose address is to be found LB = Lower limit / Lower Bound of subscript, if not specified assume 0 (zero) Example: Given the base address of an array B[1300…..1900] as 1020 and size of each element is 2 bytes in the memory. Find the address of B[1700]. Solution: The given values are: B = 1020, LB = 1300, W = 2, I = 1700 Address of A [ I ] = B + W * ( I – LB ) = 1020 + 2 * (1700 – 1300) = 1020 + 2 * 400 = 1020 + 800 = 1820 [Ans] read less

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While storing the elements of a 2-D array in memory, these are allocated contiguous memory locations. Therefore, a 2-D array must be linearized so as to enable their storage. There are two alternatives to achieve linearization: Row-Major and Column-Major. two-dimensional-array-memory-address-calculation row-major-column-major-memory-address-calculation Address... read more

While storing the elements of a 2-D array in memory, these are allocated contiguous memory locations. Therefore, a 2-D array must be linearized so as to enable their storage. There are two alternatives to achieve linearization: Row-Major and Column-Major. two-dimensional-array-memory-address-calculation row-major-column-major-memory-address-calculation Address of an element of any array say “A[ I ][ J ]” is calculated in two forms as given: (1) Row Major System (2) Column Major System Row Major System: The address of a location in Row Major System is calculated using the following formula: Address of A [ I ][ J ] = B + W * [ N * ( I – Lr ) + ( J – Lc ) ] Column Major System: The address of a location in Column Major System is calculated using the following formula: Address of A [ I ][ J ] Column Major Wise = B + W * [( I – Lr ) + M * ( J – Lc )] Where, B = Base address I = Row subscript of element whose address is to be found J = Column subscript of element whose address is to be found W = Storage Size of one element stored in the array (in byte) Lr = Lower limit of row/start row index of matrix, if not given assume 0 (zero) Lc = Lower limit of column/start column index of matrix, if not given assume 0 (zero) M = Number of row of the given matrix N = Number of column of the given matrix read less

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Absolutely, academic books do not have multidimensional arrays explained in depth. We are familiar with mostly 2X2 or 3X3 i.e 2 dimensional Arrays. lets say int rupendra ; means that 10 specifies the total number of items in a particular row, so there are 5 rows AND 5 columns too. I think its sufficient... read more

Absolutely, academic books do not have multidimensional arrays explained in depth. We are familiar with mostly 2X2 or 3X3 i.e 2 dimensional Arrays. lets say int rupendra [5][5][10]; means that 10 specifies the total number of items in a particular row, so there are 5 rows AND 5 columns too. I think its sufficient for now. Cant explained everything here. read less

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Array of an element of an array say A is calculated using the following formula: Address of A = B + W * ( I -- LB ) Where, B = Base address W = Storage Size of one element stored in the array (in byte) I = Subscript of element whose address is to be found LB = Lower limit / Lower Bound of... read more

Array of an element of an array say “A[ I ]” is calculated using the following formula: Address of A [ I ] = B + W * ( I – LB ) Where, B = Base address W = Storage Size of one element stored in the array (in byte) I = Subscript of element whose address is to be found LB = Lower limit / Lower Bound of subscript, if not specified assume 0 (zero) read less

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Given the base address of an array B as 1020 and size of each element is 2 bytes in the memory. Find the address of B. Solution: The given values are: B = 1020, LB = 1300, W = 2, I = 1700 Address of A = B + W * ( I -- LB ) = 1020 + 2 * (1700 -- 1300) = 1020 + 2 * 400 = 1020 + 800 = 1820... read more

Given the base address of an array B[1300…..1900] as 1020 and size of each element is 2 bytes in the memory. Find the address of B[1700]. Solution: The given values are: B = 1020, LB = 1300, W = 2, I = 1700 Address of A [ I ] = B + W * ( I – LB ) = 1020 + 2 * (1700 – 1300) = 1020 + 2 * 400 = 1020 + 800 = 1820 [Ans] read less

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using pointer in program

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Row Wise: The address of a location in Row Major System is calculated using the following formula: Address of A = B + W * Column Wise: The address of a location in Column Major System is calculated using the following formula: Address of A Column Major Wise = B + W * Where, B... read more

Row Wise: The address of a location in Row Major System is calculated using the following formula: Address of A [ I ][ J ] = B + W * [ N * ( I – Lr ) + ( J – Lc ) ] Column Wise: The address of a location in Column Major System is calculated using the following formula: Address of A [ I ][ J ] Column Major Wise = B + W * [( I – Lr ) + M * ( J – Lc )] Where, B = Base address I = Row subscript of element whose address is to be found J = Column subscript of element whose address is to be found W = Storage Size of one element stored in the array (in byte) Lr = Lower limit of row/start row index of matrix, if not given assume 0 (zero) Lc = Lower limit of column/start column index of matrix, if not given assume 0 (zero) M = Number of row of the given matrix N = Number of column of the given matrix read less

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