how to calculate address of n dimensional matrix
Asked by Rupendra Singh Last Modified
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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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A textbook I recently read discussed row major & column major arrays. The book primarily focused on 1 and 2 dimensional arrays but didn't really discuss 3 dimensional arrays. I'm looking for some good examples to help solidify my understanding of addressing an element within a multi-dimensional array...
read more A textbook I recently read discussed row major & column major arrays. The book primarily focused on 1 and 2 dimensional arrays but didn't really discuss 3 dimensional arrays. I'm looking for some good examples to help solidify my understanding of addressing an element within a multi-dimensional array using row major & column major arrays. 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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Address is always remain same for one or n dimensional matrix
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Satish kumar paswan
Computer
Using pointer u can calculate the address of matrix but it will be machine architecture dependent
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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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