GATE CSE 2002 | Question: 2.8

Question

Consider the following declaration of a two-dimensional array in C:

char $a[100][100]$;

Assuming that the main memory is byte-addressable and that the array is stored starting from memory address $0$, the address of $a [40][50]$ is:

• A. $4040$
• B. $4050$
• C. $5040$
• D. $5050$

Comments

Thanks got it...😊

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&a[40][50]=&*(*(a+40)+50)=*(a+40)+50

&a is a pointer to complete 2D array.

a is a pointer to first row of 2D array.

*a is a pointer to first element of first row of 2D array.

**a is the first element of 2D array.

a+1 to 2^nd row and so on….

So if starting address of array is 0 then a=0

a+1 points to next row so a+1=100 (100 char in each row)

a+2=200….

a+40=4000

now *(a+40) points to first character of 41th row

So  *(a+40)+1=4001

*(a+40)+2=4002…….

*(a+40)+50=4050.

There is no need to remember any formula :)

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NOTE:

for 2D array, A[D1][D2]

we want to know address of A[i][j]

Address = BaseAddress + (Offset x Element size)

Offset = (i x D2) + j

Answer 1

The answer is (B).

In $\mathbb{C},$ arrays are always stored in the row-major form.

Formula to evaluate $2-D$ array's location is:

 $loc(a[i][j]) = BA + [(i-lb_1)\times NC+(j-lb_2)]\times c$

where,

• $\text{BA}$ - Base Address
• $\text{NC}$ - no. of columns
• $c$ - memory size allocated to data type of array $a[lb_1 \ldots ub_1] [lb_2\ldots ub_2]$

Here, $\text{BA} = 0,  \text{NC}  = 100,    c=1,    a[0 \ldots 99][0\ldots 99]$, so $lb_1=0 , lb_2=0$

$loc(a[40][50])= 0+ [ (40-0)\times 100 + (50-0)]\times 1$

$\qquad \qquad \quad = 0+[4000+50]\times 1 = 4050$.

Comments

what are lb1 and lb2??

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i think by these ans may be contradict but by default we use  column majaor

.its a naive approach :)

Any newbie please note that this information is incorrect.

Row Major Order is default in C (unrelated: Static scoping is also default in C)

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Small thing but very img , before that i though both are right

Answer 2

$a$        $\underbrace{[100]}$            $\underbrace{[100]}$

        $\text{Streets}$       $\text{Buildings}$

Now you want to go and meet a friend who lives on $50^{th}$ building of $40^{th}$ street means $a[40][50]$

So, firstly cross $39 (0-39)$ streets each consist of $100$ buildings: $40\times 100=4000$

Now cross $49(0-49)$ buildings in order to reach your destination: $50$  

$Ans: 4050$

Comments

Good analogy! I always use "Skip i rows and j columns", thanks to you I'll add a story behind that from now on.

Elaborating your answer for completeness' sake.

 

Skip 40 rows and 50 columns

=> Skip 40*(Size of column) elements + 50 elements.

=> Skip 40*100 elements + 50 elements.

=> Skip 4050 elements.

 

Here, the element is char, and 1 char takes up 1 Byte.

So, 4050 Bytes. => 4050th address.

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Also, note that this approach works when the initial element is at A[0][0]. If somewhere else like A[-5][2], then skip accordingly.

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My man forgot about base address... And size of element, lower bound,But still not matter in this case

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great ans! this shows that there is no need to memorize any formula to calculate.

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great explanation! This shows that there is no need to memorize any formula to calculate.

Answer 3

OPTION B

Answer 4

Address(a[i][j]) = Base + (i * COL + j) * size

Step Tree:
1. size = sizeof(datatype)  --> here char = 1
2. COL = given second dimension (100)
3. Substitute i=40, j=50
4. Address = (40*100 + 50)*1 + 0 = 4050

So final = 4050

 

Answer 5

Considering a 2D array:

Answer 6

METHOD-1

METHOD-2

We don't need to rely on formulas for everything. It's just simple aptitude.
Answer: OPTION B