Programming and Data Structures Test 2

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Programming and Data Structures Test 2

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Question 1
1 / -0

Consider the following recursive function.
Int function (int x, int y) {
                If (y <= 0) return x;
                return function (y, x%y);
                }
The above recursive function computes ______.

A
y^x

B
GCD of x and y

C
x^y

D
LCM of x and y

Solution

Consider x = 10 and y = 25
CASE1:
If (y <= 0) return x;                           // 25<=0 (false)
                return function (y, x%y);            return function(25, 10)
CASE 2:
x = 25, y= 10
If (y <= 0) return x;                          // 10<=0 (False)
                return function (y, x%y);        // return function (10, 5)
CASE 3:
x =10, y =5
If (y <= 0) return x;                         // 5<=0 (false)
                return function (y, x%y);             //return function(5, 0)
CASE 4:
x = 5, y =0
If (y <= 0) return x;                         // condition true, return 5
                return function (y, x%y);
Output = 5
Which is greatest common divisor of 10 and 25.
So, given program computes GCD of x and y
Question 2
1 / -0

Consider a 1-D array in which the index starts from 2 and ends at 153. The address of index 2 is 1100 and each data in the array takes 4 words. What is the address of A[125] where 125 is the index?

A
1592-1592

B
15924-15921

C
15927-15926

D
15929-15922

Solution

Data:
base address = 1100
starting index = 2
index = 125
size of element = 4 word
Formula:
Array index starts with 0:
Address of array location : base address + index × size of data
Array index starts with 2:
Address of array location : base address + (index - 2) ×size of data
Calculation
Address of A[125] = 1100 + (125 - 2) × 4
= 1100 + 123 × 4
= 1592
Question 3
1 / -0

Consider the below given code:
#include <stdio.h>
int main()
{
char c[] = "INDIA_CAPITAL_DELHI";
char *p = c;
printf("%s", p - p[2] - p[6] + p[7] + p[1]);
return 0;
}
The number of characters in the output is _____.

A
11-11

B
114-111

C
117-116

D
119-112

Solution

Index 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19
Character
I
N
D
I
A
_
C
A
P
I
T
A
L
_
D
E
L
H
I
\0
Address 100 101 102 103 104 105 106 107 108 109 118 119
pointer
p

Let us assume ASCII value of A be x
p = 100
p[2] = D= x + 3
p[6] = C = x + 2
p[7] = A = x
p[1] = N = x + 13
p - p[2] - p[6] + p[7] + p[1] =
= 100 - (x + 3) - (x + 2) + x + (x + 13)
= 108
print the character staring from address 108 until null (\0) is encountered
Output: PITAL_DELHI
∴ number of character in the output character array is 11

Question 4

1 / -0

Consider the following belw mentioned C code:

#include <stdio.h>
int call( ) {
static int x = 13;
return x--;
}
int main( ) {
int count = 1;
for(call( ); call( ); call( ))
count = count + call();
printf("%d", count);
return 0;
}

What is the output on the execution of the program?

27-27

274-271

277-276

279-272

Solution

Initially count = 0

Iteration	Function call	Value returned by function	Explanation	count
1st iteration:	for (call( ); call( ); call( ) ) count = count + call( )	for (13; 12; fun ( ) ) count = 1 + 11	→ Initialization statement: Optional → Conditional statement = 12 and x = 11	12
2nd iteration:	for (call( ); call( ); call( )) count = count + call( )	for (13; 9; 10) sum = 12 + 8	→ Incremental statement = 10 and x = 9 → Conditional statement = 9 and x = 8	20
3rd iteration	for (call( ); call( ); call( )) count = count + call( )	for (13; 6; 7) count = 20 + 5	→ Incremental statement = 7 and x = 6 → Conditional statement = 6 and x = 5	25
4^th iteration	for (call( ); call( ); call( )) count = count + call( )	for (13; 3; 4) count = 25 + 2	→ Incremental statement = 4 and x = 3 → Conditional statement = 3 and x = 2	27
5th iteration	for (call( ); call( ); call( )) scount = count + call( )	for (13; 0; 1) count statement will not execute	→ Incremental statement = 1 and x = 1 → Conditional statement = 0 and x = -1 → Loops conditions is false	27

Output of the code is 27.

Important Points:

Initialization statement is executed only once.

Incremental statement is executed at the end of every iteration

Question 5
1 / -0

A is a 2D-array with the range [-5…..5, 3…..13] of elements. The starting location is 100. Each element occupies 2 memory cells. Calculate the location of A[0] [8] using column major order.

A
100

B
110

C
220

D
240

Solution

Concepts:
Range of given array: [-5....5,3......13]
Old location (Given): A[0][8]
A[-5][3] = 100
Number of rows: 5 – (–5) + 1= 11
Number of columns: 13 – 3 + 1 = 11
Array size: A_{11 x 11}
New range: [0 ...10, 0 ... 10]
New location: A[0 + 5][8 -3] = A[5][5]
Size of cell = 2
A[0][0] = 100
Formula:
A[n][m] = A[0][0] + (11m + n)× Size of cell
Calculation:
A[5][5] = 100 + (11× 5 + 5)× 2
∴ A[5][5] = 100 + 60 × 2
∴ A[5][5] = 220.
Old location of A[0] [8] = New location of A[5] [5] = 220.

Question 6

1 / -0

Consider the following C functions.

int f1 (int m) {

static int j = 0;

if (m > 0) {

++j;

f1(m-1);

}

return (j) ;

}

int f2 (int m) {

static int j = 0;

if (m > 0) {

j = j + f1(m) ;

f2(m-1) ;

}

return (j) ;

}

The return value of f2(4) is ______

30-30

304-301

307-306

309-302

Solution

Variable ijis static in both the functions thus its value will persist throughout the program.
When f2 (4) gets called, it will first call f1 with the value of n as 5 and then f2 will get called with n = 4.
Similarly, the same steps are taken for all the upcoming values of n.

f2(4) called

0 + fun1 (4)

f1 (4) called
f1 (3)
f1 (2)
f1 (1)
f(0) -> fun1 (0) returns 0 + 4 = 4.
f1 (1) -> 4
f1 (2) -> 4
f1 (3) -> 4
f1 (4) -> 4

Fun2 (3) called

4 + fun1 (3)

f1 (3) called
f1 (2)
f1 (1)
f1 (0) -> f1 (0) returns 4 + 3 = 7.
f1 (1) -> 7
f1 (2) -> 7
f1 (3) -> 7

f2(2) called

4 + 7 + fun1 (3)

f1 (2) called
f1 (1)
f1 (0) -> f1 (0) returns 7 + 2 = 9.
f1 (1) -> 9
f1 (2) -> 9

f2(1) called

4 + 7 + 9 + fun1 (3)

f1 (1) called
f1 (1)
f1 (0) -> f1 (0) returns 9 + 1 = 10.
fun1 (1) -> 10

f2 (0) called

f2 (0) returns 4 + 7 + 9 + 10 = 30
f2 (1) -> 30
f2 (2) -> 30
f2 (3) -> 30
f2 (4) -> 30
f2 (5) -> 30

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Index	0	1	2	3	4	5	6	7	8	9	10	11	12	13	14	15	16	17	18	19
Character	I	N	D	I	A	_	C	A	P	I	T	A	L	_	D	E	L	H	I	\0
Address	100	101	102	103	104	105	106	107	108	109									118	119
pointer	p

Programming and Data Structures Test 2

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Programming and Data Structures Test 2

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Question 1

yx

GCD of x and y

xy

LCM of x and y

Solution

Question 2

1592-1592

15924-15921

15927-15926

15929-15922

Solution

Question 3

11-11

114-111

117-116

119-112

Solution

Question 4

27-27

274-271

277-276

279-272

Solution

Question 5

100

110

220

240

Solution

Question 6

30-30

304-301

307-306

309-302

Solution

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y^x

x^y