Array Access Problem Solution

Array Access Problem Solution
	Array Access Problem

On the dope vector for

a: array[2..10, 3..8] of integer;

First, the long way:

To find the L-value for a[i,j], you must skip down i rows, then along j columns. We move along a row the distance past the beginning, that is, the distance past 3. Each step is of size 4 bytes, so we move along the row a distance of 4×(j−3). To reach the row, we must likewise skip down i rows from the start of the array. The size of a row is 4×(8−3+1), and we need to pass i−2 of them. Therefore, the L-value of the a[i,j] is

5000 + 4×(8−3+1)×(i−2) + 4×(j−3)

After algebraic surgery, that expression becomes:

4940 + 24×i + 4×j

Which produces the vector:

V₀ = 4940

d₁ = 24

d₂ = 4

The short way is to recall that the algebra has already been done. The d₂ value is always the size of the array element type, and

d₁ = (U₂ − L₂ + 1)×d₂,
V₀ = α − L₁×d₁ - L₂×d₂.

Then plug.

The vector tells you that the location of a[5, 7], for instance, is

4940 + 5×24 + 7×4 = 5088

For the slice, b = a[*,6], you need to see that b[i] is just another way to write a[i,6]. So the L-value of b[i] can be given as:

4940 + 24×i + 4×6 = 4964 + 24×i

Which can be computed from this dope vector:

V₀ = 4964

d₁ = 24

The slides give the formula

V₀ = L-value(A[L₁, j]) - d₁×L₁ = L-value(A[2, 6]) − 48 = 5012 &minus 48 = 4964

Along with d₁ for the step value, which produces the same vector.