GATE CSE 2014 | Set 2 | Question: 34

Question

For a C program accessing $\mathbf{X[i] [j] [k]}$, the following intermediate code is generated by a compiler. Assume that the size of an integer is $32$ bits and the size of a character is $8$ bits. 

t0 = i ∗ 1024 
t1 = j ∗ 32
t2 = k ∗ 4 
t3 = t1 + t0 
t4 = t3 + t2 
t5 = X[t4]

Which one of the following statements about the source code for the C program is CORRECT?

• A. $\mathbf{X}$ is declared as "int $\mathbf{X[32] [32] [8]}$”.
• B. $\mathbf{X}$ is declared as "int $\mathbf{X[4] [1024] [32]}$”.
• C. $\mathbf{X}$ is declared as "char $\mathbf{X[4] [32] [8]}$”.
• D. $\mathbf{X}$ is declared as "char $\mathbf{X[32] [16] [2]}$”.

Comments

@Kshyamano In the link that you posted , it doesnt says how to calculate no of 2d arrays in 3d array. It is option elimination after one point.

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thnk you...it was understandable

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thanks

Answer 1

$k$ is multiplied by $4$, means sizeof(dataype) is int.
$j$ is multiplied by $32$, means the inner most dimension of array is $32/4 = 8$ (we have to divide by the size of the inner dimension- which here is a simple integer)
$i$ is multiplied by $1024$, means the second dimension of array is $1024/32 = 32$ ($32 = 8*4$ is the size of the inner dimension here)

So, (A) is correct. The first dimension is not needed for code generation and that is why in C language while passing an array to a function, we can omit the value of the first dimension but not any others.

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We can also do as follows:

$X[i][j][k] = *(*(*(X + i) + j) + k)$

In Integer arithmetic, this equals

$*(*(*(X + i * sizeof(*X) ) + j * sizeof(**X) + k * sizeof(***X) )$

as for every add to a pointer we have to multiply the size of the pointed value (to get a valid address)

So, from the given code we get

$sizeof(***X) = 4, -$ int
$sizeof(**X) = 32 -$ int  array  of size $8$
$sizeof(*X) = 1024 - 2$ $D$ int array of size $[32]$ havinf size of inner $1D$ array $32$.

So, the inner dimensions must be $32$ and $8$ and type must be integer. So, only option A matches.

Comments

Any Reference link for this Answer?

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Can anyone elaborate further did not understand

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How is r1 calculated ?

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See now..

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@Das if u mean dimension one - that cant be calculated.. only assumed from choice..

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Thank you Sir :)

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Sir can u explain why we concluded that since k is multiplied by 4 the data type is int ????

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Difference between option A) and C)?

Why C) cannot be answer?

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Exactly... why c is wrong? It also has its outer dimensions 32 and 8 only

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@srestha 

why c is wrong? It also has its outer dimensions 32 and 8 only

This is because 

Inner dimensions must be 32 and 8 and type must be integer

And option C is char type..so wrong.

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@Arjun Suresh

Why only k is multiplied by 4 (size of each element), If 4 is size of each element of the array, then why multiply k alone, why not all the remaining elements in the 2D and 1D arrays (1024*i+32*j number of elements) we have surpassed to reach the desired element.

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why the last element is 8. please explain the logic

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Suppose given dimensions are X[a][b][c] and you have to find element at X[i][j][k] It will be done like

(i * b * c * size) + (j * c * size) + (k *size). Now in the code k is multiplied by 4 so we got size as 4. Now go in reverse way to calculate c (innermost dimension) we have 

t1 = j ∗ 32 ----1

Our second term (j * c * size) ----2

c*size=32 and we know size is 4 so c is 8.

same we get b as 32.

Hope you got it. :)

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For those asking about why it's an integer array, from the second explanation @Arjun da has given, since $sizeof(***X) = 4$, and $sizeof$ returns the size of a type in bytes, 4 bytes = 32 bits = integer type.

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@Arjun

@Vikrant Singh

why we are access array like this,

actually 3d array address resolution  should be like,

FORMULA

**Row Major and Column Major for 3-D Array:
**ROW MAJOR OF A[I, J, K]=B+W[(K-Ko)*RC+(I-Io)*C+(J-Jo)] 
COLUMN MAJOR OF A[I, J, K]=B+W[(K-Ko)*RC+(I-Io)+(J-Jo)*R]**

WHERE:
R=ROW
C=Column
K=Weidth
Ko=Lower Bound of Width
Io=Lower Bound of Row
Jo=Lower Bound of Column**

https://stackoverflow.com/questions/33463594/element-address-in-3-dimensional-array

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$∗(∗(∗(X+i∗sizeof(∗X))+j∗sizeof(∗∗X){\color{Blue} )}+k∗sizeof(∗∗∗X))$

one closing bracket is missing :P

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Awesome explanation,cleared my concepts regarding 3d array.

Answer 2

X is of type integer and dimensions of X are r₁ X r₂ X r₃.

 value of r₂ = 32

r₃ = 8

Comments

So, if here t2= k*1 instead of k*4 then answer would be for Char array, is it?

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it would be char array if we multiplied by 1 instead of 4 ( as character takes 8bits - 1 byte)

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K = A[2]; Lets write intermediate code for this.

A[j]=*(Base Address + j*intsize)

A[j] = *(A + j*intsize) = *( 1000 + 2*4) = *(1008) = 30 ( our element)  

i have doubt in this ..the arrays starting index is 0 so if you are taking A[2] then you have to like that 1000 + (2-0+1)*4

Answer 3

For a while, assume that this 3-D array is of int type.

If I write int k = X[i][j][k]; Means I am looking for a particular integer value which is stored at the location *(X[i][j]+k) = *(*(X[i]+j)+k) = *(*(*(X+i)+j)+k)

Lets take an example of 1D array

 if there is one 1D array A[10,20,30,40], Suppose, A is stored at 1000 location, intsize=4B 
and If I write int k = A[2], means i am looking for an integer which is stored at *(A+2) location.

K = A[2]; Lets write intermediate code for this.

A[j]=*(Base Address + j*intsize)

A[j] = *(A + j*intsize) = *( 1000 + 2*4) = *(1008) = 30 ( our element)

t1 = j*intsize
t2 = baseAddress + t1
k = *(t2)  it will give 30 to us.

Now, take an example of 2D array.

Int k = B[3][5]; It will also assign k with an integer value. We have to find the location of that integer.

Suppose array B has x rows and y columns, and stored in memory in ROW MAJOR Order.

k = B[i][j] = *(B +([i*no.Of_Cols]+j)*intsize)
k = B[3][5] = *(B + ((3*y) + 5)*4)
Suppose number of rows = 30, number of columns =20
k=B[3][4]= *(Base address + ((3*20)+5)*4) = *(B.A + 3*20*4 + 5*4) = *(1000 + 240+ 20) = *(1260).
To access, B[3][5], i can also write B[260/4] = B[65] = *(B + 65*4) = *(1000 + 260) = *(1260)
Intermediate code for k = B[i][j];

t1 = j*intsize     = 5*4 = 20
t2 = i*#cols*instize  = 3*20*4 = 240
t3 = t1+t2  = 260
k = B[t3]; *(1000 + 260) = *(1260);

PS: Assume Array B is stored at memory location 1000.

Let's take 3-D array now.

int k = C[i][j][k].

It means we have to cross, i 2D arrays, j 1D arrays, and k elements to reach our required element.

= B.A + (i_2D_arrays*no._elements_In_1_2D_Array + j*no._elements_in_1_1D_Array + k)*intsize

=B.A + (i_2D_arrays*no._elements_In_1_2D_Array*instsize + j*no._elements_in_1_1D_Array*intsize + k*intsize)

Now, we can see in question that k is multiplied by 4, hence array X is of int type. options (C) and (D) are eliminated here.

j is multiplied by 32, and this calculation equals to j*no._elements_in_1_1D_Array*intsize=j*32 = j*8*4
it means there are 8 elements in one 1D array.

int X[][][8];  so far so good to eliminate option B.

Let's go further.

i*1024 = i*no._elements_In_1_2D_Array*intsize

number of columns in 2D arrays = 8
number of rows in 2D arrays 256/8 = 32

Hence, there 2D array is of [32][8] size.

i*1024 = i*32*8*4 = i is number of 2D arrays, 32x8 is size of one D array. 

We can't calculate number of 2D arrays by the given code, and that is not even required for our question.

So, the answer is (A). 

Comments

@Pranavpurkar yes in that codition answer will be both A and C….

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How do we know it is row major

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Really a nice explanation , thanks a lot !.

Answer 4

$int\ X$	$\underbrace{32}$	$\underbrace{32}$	$\underbrace{8}$
$Assume$	$streets$	$apartments$	$buildings$

According to question we need to find: $loc\ X[i][j][k]$

Size of int$:4B$

$Analogy:$ You want to go and meet your friend who lives in $k^{th}$ building of $j^{th}$ apartment which is on $i^{th}$ street.

1. First you will cross $i$ streets which contains $32$ apartments and in each apartment there are $8$ buildings

$\therefore\ 32\times 8\times 4=1024$

Hence $t_{0}=i\times 1024$ 

2. Now you will cross $j$ apartments which contains $8$ buildings

$\therefore\  8\times 4=32$

Hence $t_{1}=j\times 32$ 

3. Now you will cross $k$ buildings

Hence $t_{2}=k\times 4$ 

Rest of the part everyone knows :)

Comments

Hmmm.. kiran sir haan! :)

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taking the answer and stress testing Lmao

Answer 5

The final expression can be simplified in form ofi, j and k by following the intermediate code steps in reverse order

t5 = X[t4]
   = X[t3 + t2]
   = X[t1 + t0 + t2]
   = X[i*1024 + j*32 + k*4]
   = X + i*1024 + j*32 + k*4 

Since k is multiplied by 4, the array must be an int array. We are left with 2 choices (A and B) among the 4 given choices. X[i][j][k]'th element in one dimensional array is equivalent to X[i*M*L + j*L + k]'th element in one dimensional array (Note that multi-dimensional arrays are stored in row major order in C). So we get following equations

j*L*4 = j*32, we get L = 8 (4 is the sizeof(int))
i*1024 = i*M*L*4, we get M = 1024/32 = 32

Therefore option A is the only correct option as M and L are 32 and 8 respectively only in option A.

Comments

It's size of data type mentioned in question.i.e., 32 and 8

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@Regina Phalange Ma'am why we are multiply j with L  like j*L and for i with M*L

Also how are you evaluating this   X + i*1024 + j*32 + k*4 to X[i*M*L + j*L + k]?

from where did this X[i*M*L + j*L + k] came?

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This is answer for me. Thanks

If given permission , I would have up voted answer at least 10 times

Answer 6

Answer - A

Explanation - 

Element needed - X[i][j][k]

A 3D array is a collection of 2D arrays and a 2D array is a collection of 1D arrays, and a 1D array is a collection of elements.

So, we need to first cross the required no. of 2D arrays(no. of elements in 1 2D array * i), then the required no. of 1D arrays(no. of elements in 1 1D array(no. of rows in 1 2D array) * j) in our reached 2D array, and then the required no. of elements in our reached 1D array(k).

1. We have t0 = i * 1024, which means that there are 1024 elements in 1 2D array.

2. We have t1 = j * 32, which means that we have 32 elements in 1 1D array of our 2D arrays or in other words, the no. of rows of our 2D array is 32, so the no. of columns can easily be found by 1024 / 32 = 32. 

So now we have the dimensions of our 2D arrays = 32*32.

so clearly our answer is A (no other close option).

elaborating further, 

3. We have calculated t3 of our 3-address code = (i * (32*32)) + (j * 32) + k.

4.We now need to multiply the above expression with the size of the datatype. 

Since we have t2 = k * 4, it clearly means that the size of our datatype of each element is 4 bytes, hence an int. 

And the expression t4 of our 3-address code in high level language is - ((i * (32*32)) + (j * 32) + k) * 4

5. t5 gives the specific element. 

Conclusion - our required 3D array is - x[32][32][8] and it is of type integer. 

Note - There is no way of finding out the 3rd dimension of our 3D array, here 8 with the help of our 3-address code, since its not required in the calculation of an element's location.

Comments

@ ravi_ssj4

Sir, I think u have made a mistake in point 2. j is the number of rows and 32 is the number of elements in each row, although this should not effect final answer.

And I could not understand why they have multiplied k by 4. If it is size of each element of the array, then why multiply k alone, why not all the remaining elements in the 2D and 1D arrays we have surpassed to reach the desired element.