UGC NET CSE | December 2014 | Part 2 | Question: 24

Question

Consider an array $A\left[20, 10\right]$, assume $4$ words per memory cell and the base address of array $A$ is $100$. What is the address of $A\left[11, 5\right]$ ? Assume row major storage.

• A. $560$
• B. $565$
• C. $570$
• D. $575$ 

Answer 1

Each cell needs 4 words

• A[0][0], A[0] [1],.....A[0][9] needs 4x10 =40 words.
• Since array starts at 100, A[0][9] will be at 100+40=140 th location.
• Similarly, from A[0][0] to A[10][9] there are 11x10=110 elements.
• From A[11][0] to A[11][5]there are 5 elements.
• Total 115 elements are present before A[11][5].
• Hence A[11][5] will be at 100+ 4x115 = 560

Answer is A

Comments

We can simply use the row-major order formula to find the array address of A[I,J].

Let A[M,N] be the size of the array, w be the word size, then

A[I,J] = Base Address of A + w(I *N + J)

Here, A[M,N] is A[20,10]

So, A[11,5] = 100 + 4(11*10 + 5)

                      = 100 + 4(115) = 100+460

                                    =560

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How the number of columns is 10 and not 11??

0 being 1^st column, 10 should be 11^th column. Isn’t it?

Number of column given in the matrix N = Upper Bound – Lower Bound + 1 = 10 -0 +1 =11 

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Why the lower bound of row is 1 instead of 0? means it should be 11*11+5?

Answer 2

row and coloum major

Answer 3

A[20][10]

to reach A[11][5]

we have to cross 11 rows and each row contains 10(no. of columns) elements and in 12^th row we have to cross 5 elements 

total elements to cross= $11*10+5=115$

each element occupies 4 words space 

total space = $4*115=460$

address of $A[11][5] = base address + space occupied by elements before it$

                                $=100+460=560$

option (A)