Question - 1 : -
Observe the following C++ code and find out, which out of the given options (i) to (iv) are the expected correct output. Also assign the maximum and minimum value that can be assigned to the variable ‘Go’.

void main()
{ int X [4] = {100, 75, 10, 125};
int Go = random(2) + 2;
for (int i = Go; i<4; i ++)

cout<
}
(i) 100$$75
(ii) 75$$10$$125$$
(iii) 75$$10$$
(iv) 10$$125$

Answer - 1 : -

(iv) is correct option

Max. Value = 3

Min. Value = 2

Question - 2 : -
Write the difintion of a function grace_score (int score [], int size) in C ++,which should check all the elements of the array and give an increase of 5 to those scores which are less than 40.
Example: if an array of seven integers is as follows:
45,35,85,80,33,27,90
After executing the function, the array content should be changed as follows :
45,40,85,80,38,32,90

Answer - 2 : -

void grace_score(int score[].int size)

{

for(int i=0;i

{

if (score [i]<40)

score [i]=score[i]+5;

cout< }

}

Question - 3 : -
Find the output of the following C+ + program:

#include void repch(char s[]) { for(int i=0;s[i]!=’\0′;i++) { if(((i%2)!=0) &&(s[i]!=s[i + 1])) { S[i]=’@’; cout<<“Hello”; } else if (s[i]==s [i + 1]) { s[i+1]=” ! ” ; i++ ; } } } void main() { char str[]=”SUCCESS”; cout<<“Original String”<

Answer - 3 : -

Original String SUCCESS

Changed String S@C!ES!

Question - 4 : -
Find the output of the following:

#include
void switchover (int A[],int N, int split)
{

for(int K=0; K

if(K
A [K] + = K;
else
A[K] *=K; }
void display(int A [],int N)
{

for (in K=0; K

(K%2==0)? Cout<

cout<
}
void main()
{ int H [ ] = {30, 40, 50, 20, 10, 5};
switchover(H, 6, 3);
display(H, 6);
}

Answer - 4 : -

30%41

52% 60

40% 25

Question - 5 : -
An integer array A [30] [40] is stored along the column in the memory. If the element A[20][25] is stored at 50000, find out the location of A[25] [30].

Answer - 5 : -

A[i] [j] = B+W x [No. of rows x (I-Lr) + (J-Lc)]

A[20] [25] = B + 2x [30x (20-0) + (25-0)]

50000 = B+2x[30x(20-0)+(25-0)]

B = 48750

A[7] [10] = 48750+ 2x[30x(7-0)+(10-0)]

= 49190

Question - 6 : -
An array P[30][20] is stored along the column in the memory with each element requiring 2 bytes of storage. If the base address of the array P is 26500, find out the location of P[20][10].

Answer - 6 : -

Total number of rows = 30

Total size = 2 bytes

Base Address = 26500

LOC(P[I][J]) = Base Address+((I-LBR)+(J- LBC)*R)W

Assuming Lower Bound of Row(LBR)=0

Lower Bound of Column(LBC)=0

Total number of Rows(R)=30

Size of each element(W)=2

LOC(P[20][10]) = 26500 + ((20-0)+(10-0)*30)*2

LOC(P[20][10]) = 26500 + 640

LOC(P[20][10]) = 27140

Question - 7 : -
Write a function to sort any array of n elements using insertion sort. Array should be passed as argument to the function.

Answer - 7 : -

void insertsort (int a[],int n)

{

int p,ptr;

//Assuming a[0]=int_min i.e. smallest integer

for(p = 1; p<=n;p++)

{

temp=a[p];

ptr=p-1;

while (temp

{

a [ptr+1]=a[ptr] ; //Move Element

Forward

ptr--;

}

a[ptr+1]=temp; //Insert Element in Proper Place

}

Question - 8 : -
Write a function NewMAT(int A[][],int r, int c) in C++, which accepts a 2d array of integer and its size as parameters divide all those array elements by 6 which are not in the range 60 to 600(both values inclusive) in the 2d Array.

Answer - 8 : -

void NewMAT(int A[] [] , int r,int c)

{

for (int i = 0; i

for (j=0; j=60)&&(A[1] [j)<=600))

A[i][j]/=6

A[i] [j] =A[i] [j] /6;

}

Question - 9 : -
A two dimensional array P[20] [50] is stored in the memory along the row with each of its element occupying 4 bytes, find the address of the element P[10] [30], if the element P[5] [5] is stored at the memory location 15000.

Answer - 9 : -

Loc (P [I] [J] along the row =BaseAddress+W [(I-LBR)*C+ (J-LBC)]

(where C is the number of columns, LBR = LBC =0)

LOC (P [5] [5]) = BaseAddress + W* [I*C + J]

15000 = BaseAddress + 4 * [5 * 50 + 5]

= BaseAddress + 4 * [250 + 5]

= BaseAddress + 4*255 = BaseAddress + 1020

BaseAddress = 15000 -1020 = 13980

LOC (P [10] [30]) = 13980 + 4* [10 * 50 + 30]

= 13980 + 4 * 530

= 13980 + 2120

= 16100

OR LOC (P [10] [30])

= Loc (P[5] [5]) + W[ (I-LBR) * C+ (J-LBC)]

= 15000 + 4 [(10 – 5) * 50 + (30 – 5)]

= 15000 + 4 [5 * 50 + 25]

= 15000 + 4 *275

= 15000 + 1100

= 16100

(where C is the number of columns and LBR = LBC = 1)

LOC (P [5] [5])

15000 = BaseAddress + W [(I-1) *C + (J-1)]

= BaseAddress + 4 [4*50 + 4]

= BaseAddress + 4 [200 + 4]

= BaseAddress + 4 * 204

= BaseAddress + 816

BaseAddress = 15000 – 816 = 14184

LOC (P [10] [30])

= 14184 + 4 [(10 -1) * 50 + (30 -1)]

= 14184 + 4 [9*50 + 29]

= 14184 + 4*479

= 14184 + 1916

= 16100

Question - 10 : -
Write a user-defined function DisTen (int A[] [4], int N, int M) in C++ to find and display all the numbers, which are divisible by 10. For example if the content of array is :

12 20 13
2 10 30
The output should be
20 10 30

Answer - 10 : -

void DisTen (int A[] [4] , int N, int M)

{

int i, j;

for (i=0; i

{

for (j = 0 ; j

{

if (A[i][j]%10==0)

cout << A [i] [ j] ;

}

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