GATE CSE 1998 | Question: 2.14

Question

Let $A$ be a two dimensional array declared as follows:

A: array [1 …. 10] [1 ….. 15] of integer;

Assuming that each integer takes one memory location, the array is stored in row-major order and the first element of the array is stored at location $100$, what is the address of the element $A[i][j]$?

• A. $15i+j+84$

• B. $15j+i+84$

• C. $10i+j+89$

• D. $10j+i+89$

Comments

@Lakshman Patel RJIT

Are you sure that the column-major is $\mathbf{(D)}$?

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@Satbir

Can you please verify that?

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Yes, it is $\mathbf{D}$ only for Column-Major.

Answer 1

$a$        $\underbrace{[10]}$            $\underbrace{[15]}$

        $streets$       $buildings$

Now you want to go and meet a friend who lives on $j^{th}$ building of $i^{th}$ street means $a[i][j]$

So, firstly cross $(i-1)$ streets each consist of $15$ buildings: $(i-1)\times 15=15i-15$

Now cross $(j-1)$ buildings in order to reach your destination: $j-1$  

$Ans: RMO: 100+15i-15+j-1=>15i+j+84$

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$a$        $\underbrace{[10]}$            $\underbrace{[15]}$

    $buildings$       $streets$

Now you want to go and meet a friend who lives on $i^{th}$ building of $j^{th}$ street means $a[i][j]$

So, firstly cross $(j-1)$ streets each consist of $10$ buildings: $(j-1)\times 10=10j-10$

Now cross $(i-1)$ buildings in order to reach your destination: $i-1$  

$Ans: CMO: 100+10j-10+i-1=>10j+i+89$

Comments

Nice!!

Answer 2

When the initial element of the array is at A[0][0] to reach A[i][j] we skip i rows and j columns.

But here, the initial element is at A[1][1]. So, we'll skip (i-1) rows and (j-1) columns.

 

$100+(i-1)*15+(j-1)$

=> $100+15i-15+j-1$

=> $84+15i+j$

 

Option A

Answer 3

To find address of any element in an array

base address + (no. of elements before the elements whose address we want to find)* size of each element

here array [1 …. 10] [1 ….. 15]

we want to find address of $A[i][j]$ considering row major order

to reach  i^th row we have to cross  $i-1$ rows as starting index is 1

no. of elements = (i-1)*elements in each row(=no.of columns)

                          =$i-1*15$

                          =$15i-15$

to reach j^th element in i^th row we have to cross j-1 elements in i^th  row 

total elements to cross to reach $A[i][j]=15i-15+j-1$

address of $A[i][j]$ = $100-16+15i-j$

                             =$15i+j+84$

OPTION A

 

Answer 4

(A.)  100 +[(i-1)*15 + (j-1)]*1

     = 100+[15i –15+j-1]*1

    = 100+15i-16+j

       15i+j+84 .