Kamal

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

An n dimensional matrix can be of any dimension. Adding a dimension is adding one more index number (to access the element). e.g 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