Calculating the address of an element in an N-dimensional array

N-Dimensional Arrays: The N-Dimensional array is basically an array of arrays. As 1-D arrays are identified as a single index, 2-D arrays are identified using two indices, similarly, N-Dimensional arrays are identified using N indices. A multi-dimensional array is declared as follows:

int NDA[S₁][S₂][S₃]……..[S_N];

Explanation:

• Here, NDA is the name of the N-Dimensional array. It can be any valid identifier name.
• In the above syntax, S₁, S₂, S₃……S_N denotes the max sizes of the N dimensions.
• The lower bounds are assumed to be zeroes for all the dimensions.
• The above array is declared as an integer array. It can be any valid data type other than an integer as well.

Address Calculation of N-Dimensional Arrays:

Assumptions:

• The N-Dimensional array NDA with the maximum sizes of N dimensions are S₁, S₂, S₃, ………, S_N.
• The element whose address needs to be calculated has indices l₁, l₂, l₃, …………..l_N respectively.
• It is possible that the array indices do not have the lower bound as zero. For example, consider the following array T: T[-5…5][2……9][14…54][-9…-2].

Explanation:

• In the above array T, the lower bounds of indices are not zeroes.
• So that the sizes of the indices of the array are different now and can be calculated by using the formula:

UpperBound – LowerBound +1

• So, here the S₁ = 5 – (-5) + 1 = 11. Similarly, S₂ = 8, S₃ = 41 and S₄ = 8.

For address calculation, the lower bounds are t₁, t₂, t₃…….t_N. There exist two ways to store the array elements:

• Row Major
• Column Major

The Column Major formula:

Address of NDA[I₁][I₂]. . . [I_N] = BAd + W*[((…E_NS_N-1+ E_N-1 )S_N-2 +… E₃ )S₂+ E₂ )S₁ +E₁]

• BAd represents the base address of the array.
• W is Storage Size of one element stored in the array (in byte).
• Also, E_i is given by E_i  = l_i – t_i, where l_i and t_i are the calculated indexes (indices of array element which needs to be determined) and lower bounds respectively.

The Row Major formula:

Address of NDA[I₁][I₂]. . . .[l_N] = BAd + W*[((E₁S₂ + E₂ )S₃ +E₃ )S₄ ….. + E_N-1 )S_N + E_N]

The simplest way to learn the formulas:

For row-major: If width = 5, the interior sequence is E₁S₂ + E₂S₃ + E₃S₄ + E₄S₅ + E₅ and if width = 3, the interior sequence is E₁S₂ + E₂S₃ + E₃. Figure out the pattern in the order and follow four basic steps for the formula of any width:

• Write the interior sequence.
• Put the closing bracket after each E except the first term. So for width = 5, it becomes

E₁S₂ + E₂)S₃ + E₃)S₄ + E₄)S₅ + E₅).

• Put all the Opening brackets initially.

((((E₁S₂ + E₂)S₃ + E₃)S₄ + E₄)S₅ + E₅).

• Incorporate the base address and width in the formula.

The approach is similar for the Column Major but the pattern of the interior sequence is reverse of the row-major pattern. 

For column-major: If width =5, the interior sequence is E₅S₄ + E₄S₃ + E₃S₂+ E₂S₁ + E₁.

Example: Let’s take a multi-dimensional array A[10][20][30][40] with the base address 1200. The task is to find the address of element A[1][3][5][6].

Here, BAd = 1200 and width = 4.
S1 = 10, S2 = 20, S3 = 30, S4 = 40

Since the lower bounds are not given, so lower bounds are assumed to be zero.
E1 = 1 – 0 = 1; 
E2 = 3 – 0 = 3; 
E3 = 5 – 0 = 5;
E4 = 6 – 0 = 6.

Any of the techniques (Row-major or column-major) can be used for calculating the answer (unless specified). 
By applying the formula for row-major, directly write the formula as:

A[1][3][5][6] = 1200 + 4(((1 × 20 + 3)30 +5)40 + 6) 

 =1200 +4((23 × 30 +5)40 +6) 

=1200 + 4(695 × 40 + 6) 

=1200 + (4 × 27806) 

=112424.