GATE CSE 2015 | Set 3 | Question: 30

Question

Consider the following two C code segments. $Y$ and $X$ are one and two dimensional arrays of size $n$ and $ n \times n$ respectively, where $2 \leq n \leq 10$. Assume that in both code segments, elements of $Y$ are initialized to $0$ and each element $X[i][j]$ of array $X$ is initialized to $i+j$. Further assume that when stored in main memory all elements of $X$ are in same main memory page frame.

Code segment $1:$

// initialize elements of Y to 0
// initialize elements of X[i][j] of X to i+j
for (i=0; i<n; i++)
    Y[i] += X[0][i];

Code segment $2:$

// initialize elements of Y to 0
// initialize elements of X[i][j] of X to i+j
for (i=0; i<n; i++)
    Y[i] += X[i][0];

Which of the following statements is/are correct?

S1: Final contents of array $Y$ will be same in both code segments

S2: Elements of array $X$ accessed inside the for loop shown in code segment $1$ are contiguous in main memory

S3: Elements of array $X$ accessed inside the for loop shown in code segment $2$ are contiguous in main memory

• A. Only S2 is correct
• B. Only S3 is correct
• C. Only S1 and S2 are correct
• D. Only S1 and S3 are correct

Comments

what's the point of last line in the question .

" further assume that when stored in main memory all the elements of X are in same main memory page frame "

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@Srken read this comment  

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We can only access the first row and first column and for array X they are defined symmetrically.

So final elements in both code segments of array Y is same.

And array x elements in. Code segment 1 are in row major order therefore contiguous.

Answer 1

Option is C. Only $S1$ and $S2$ are correct because $Y$ have same element in both code and in code $1.$

Y[i] += X[0][i];

This row major order (In C, arrays are stored in row-major order)  which gives address of each element in sequential order$(1,2,3,\ldots,n)$ means we cross single element each time to move next shows  contiguous in main memory but in code $2$ for:

Y[i] += X[i][0];

We are crossing n element (row crossing with n element )to move next.

Comments

yess correct

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c,c++,python and java stores 2d array in row-major format and fortran matlab uses column major format :)

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because row major order is bydefault taken

Answer 2

Actually Answer depends upon storage scheme of array(Row major or Column major). However if it is not mentioned in the Question that which Scheme is used then assume it is stored in Row major because by default it is stored in Row major so here answer restrict to option C only.

Comments

In s3 Elements of array X accessed inside the for loop shown in code segment 2 are not contiguous in main memory  because they stored in column major order 

Answer 3

answer will be c

as s1 is right(final contents are same ) and s2 as we are getting

consider size of 4

y[0]=0(same as x[0][0])

y[1]=2(as x[0][1]=0+1=1 and y[1]=0+1)

y[2]=1(x[0][2]=0+2=2 and y[2]=0+2)

thus following row major order as in c

Answer 4

😊

Answer 5

Lets take an example, and understand the core concept of this question.

In Code Segment1:

          Y[i] += X[0][i];

Y[i] is storing the elements of X[0][i] (in row major order). so its elements will be in contiguous manner.

 

and, In Code Segment2:

           Y[i] += X[i][0];

 

Y[i] is taking the elements of X[i][0] (in column major), so its elements will be sparse, I mean, not in contiguous fashion. 

Answer 6

here Y-> 1D array n size

X-> 2D array nxn matrix

all elements of Y array are initialized to 0

element of X[i][j] = i+j

 

assume n = 3

solve code segment 1 ,

output u get is 0 1 2

solve code segment 2

output u get is 0 1 2

 

s1 is correct

so A and B option are eliminated

since its a c language , 2d array elements are stored contigous means in consecutive manner (RMO ie row major order in our memory

if u check code segment 1 , X[0][0] X[0][1] X[0][2] all the 3 elements are row

so option C is correct