Programming and Data Structures Test 1

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Programming and Data Structures Test 1

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Weekly Quiz Competition

Question 1
1 / -0

What is the output of the below given code?
#include <stdio.h>
int main()
{
int j;
j = printf("Testbook TestSeries\n");
j = printf("%d\n",j);
printf("%d\n",j);
}

A
Testbook TestSeries
20
3

B
Testbook TestSeries
19
2

C
Testbook TestSeries
1
1

D
Error Cannot assign prinft to integer

Solution

Code:
#include <stdio.h>
int main()
{
int j;
j = printf("Testbook TestSeries\n"); \\print Testbook TestSeries and return 20 to j
j = printf("%d\n",j); \\print 20 and return 3 to j
printf("%d\n",j);
}
Explanation:
printf return length of the string
Since = is right to left associative
Testbook TestSeries (will be printed) number of character 20 is returned
value of j is printed that is 20 and 3 is return
at the end value of j is printed that is 3.
Important Point:
int j = printf("%d\n",j) \\ since value of j = 20
int j = printf("20\n",j) \\assume length = 3 (2, 0, \n)
3 is returned to j
printf("%d\n",j); \\ hence 3 is printed

Question 2

1 / -0

Match the following list:

List – I (Storage class)	List – II (Storage, initial – value, scope)
(a) auto	(i) register, garbage, local
(b) register	(ii) memory, zero, global
(c) static	(iii) memory, garbage, local
(d) extern	(iv) memory, zero, local

(a) – (iv), (b) – (ii), (c) – (iii), (d) – (i)

(a) – (i), (b) – (iv), (c) – (ii), (d) – (iii)

(a) – (ii), (b) – (iii), (c) – (i), (d) – (iv)

(a) – (iii), (b) – (i), (c) – (iv), (d) – (ii)

Solution

Concept:

Storage classes are used to fully define a variable. Storage class tells us that where the variable would be stored, what will be initial value of variable (if initial value is not specifically assigned then default value), what would be the scope of variable and life of variable.

Explanation:

Four types of storage classes: auto, register, static and extern.

auto:

Features of auto storage class are :

Storage is memory
Default value is garbage
Scope is local to the block in which variable is defined
Life till the control remains within the block in which variable is defined.

register:

Features of register storage class are:

Storage is in CPU registers
Default initial value is garbage value
Scope is local to the block in which variable is defined
Life: till the control remains within the block in which variable is defined

static:

Features of static storage class are:

Storage is memory
Default initial value is zero
Scope is local to the block in which variable is defined
Life: value of variable persists between different function calls

extern:

Features of extern storage class are:

Storage is memory.
Default initial value is zero
Scope is global
Life: as long as the program’s execution does not come to an end

Question 3
1 / -0

What is the output of the below given code?
#include <stdio.h>
int main()
{
int a = 2;
int year = 2020;
char flag = 'a' + a;
switch(flag)
{
case 'a': printf("GATE CS %d ",year);
case 'b':
case 'c': printf("GATE CS %d ",year + 1);
case 'd': break;
default: printf("GATE CS %d",year + 2);
}
return 0;
}

A
GATE CS 2021

B
GATE CS 2022

C
GATE CS 2020 GATE CS 2021

D
It will not produce any output

Solution

int year = 2020;
char flag = 'a' + a = 'a' + 2 = 'c'
switch('c')
control will come to case 'c'
In case 'c':
Statement printf("GATE CS %d ",year + 1); is executed
Output: GATE CS 2021
since case 'c': doesn't have break statement it will go to case 'd'
In case 'd' break will execute and the switch will terminate.
Therefore, the output is GATE CS 2021.
Question 4
1 / -0

Consider the below given code in which integer is of 2 bytes?
#include <stdio.h>
void function(int *co, int *vid)
{
co = vid;
*co = 19;
}
int main()
{
int p = 20, q = 19;
function(&q,&p);
printf("Co-vid %d%d", p, q);
return 0;
}
The output is Co-vid ____? (Fill in the blanks)

A
1919-1919

B
19194-19191

C
19197-19196

D
19199-19192

Solution

Before function execution:
Variable
p
q
value stored
20
19
Assume address
1000
1002
At the time of function call:
Pointer Variable
co
vid
address stored
1002
1000

After function execution:
Pointer Variable
co
vid
address stored
1000
1000

*co = 19
∴ p = 19
Variable q remained unchanged
Variable
p
q
Value
19
19

The output is Co-vid 1919.
Question 5
1 / -0

Which one of the following choices given below would be printed when the following program is executed?
#include<stdio.h>
void display(int *tb[]);
int main()
{
int array1[] = {2,5,8,12,6,9};
int array2[] = {-12, 4, 6, -13,7};
int array3[] = {-8, -6, -7, 9};
int array4[] = {21, 23, 27,15,0};
int *t[] = { array1, array2, array3, array4};
display(t);
}
void display(int *tb[])
{
printf("%d and ", *(tb[2]+1));
printf("%d", tb[3][1]);
}

A
-6 and address of 2^nd element in array3

B
-5 and 23

C
-6 and 23

D
-5 and address of 2^nd element in array4

Solution

Size of a character is 1 byte:
Assume the starting addresses of array
array1
2
5
8
12
6
9
\0
Address
101
102
103
104
105
106
107
array2
-12
4
6
-13
7
\0

Address
201
202
203
204
205
206

array3
-8
-6
-7
9
\0

Address
301
302
303
304
304

array4
21
23
27
15
0
\0

Address
401
402
403
404
405
406

Assume the addresses of t:
Address
value
701
101
702
201
703
301
704
401

Assume the addresses of t:
Address
value
900
701

*(tb[2]+1) = *(*(tb + 2) + 1) = *(*(701 + 2) + 1) = *(301 + 1) = *(302) = -6
tb[3][1 = *(*(tb +3) + 1) = *(*(701 + 3) + 1) = *(401 + 1) = *(402) = 23
Important Points:
Since size of a character is 1 byte it is not included in address calculation
Question 6
1 / -0

If ‘a’ is declared as one-dimensional array in ‘C’ then
a) *(a + i) is same as *(&a[i])
b) *(a + i) is same as *a + i
c) &a[i] is same as a + i - 1
d) *(a + i) is same as a[i]
Which of the above statements are incorrect?

A
(a) and (b)

B
(a) and (d)

C
(b) and (c)

D
(c) and (d)

Solution

Assume
Decimal address
100
102
104
106
Integer
array
a[0]
a[1]
a[2]
a[3]
corresponding value
2
4
6
8
Take, i = 1 and size of interger be 2 bytes
*(a + i) = *(100 + i × size of integer)
= *(100 + 1 × 2) = *(102) = 4
*(&a[2]) = *(102) = 4
Therefore, *(a + i) is same as *(&a[i])
*a + i = *(100) + 1 = 2 + 1 = 3
Therefore, *(a + i) is not same as *a + i
&a[2] = 102
a + i - 1 = 100 + 1 - 1 = 100
Therefore, &a[i] is not same as a + i - 1
a[i] = a[1] = 4
Therefore, *(a + i) is same as a[i]
Hence statement b and c are incorrect
Code to verify the output:
#include<stdio.h>
int main( )
{
int i = 1;
int a[] = {2, 4, 6, 8};
printf("%d = %d\n",*(a + i),*(&a[i])); // 4 = 4
printf("%d = %d\n",*(a + i), *a + i); // 4 = 3 (incorrect)
printf("%d = %d\n",&a[i],a + i - 1); //address of 2nd element = address of 1st element (incorrect)
printf("%d = %d\n",*(a + i),a[i]); // 4 = 4
return 0 ;
}

Question 7

1 / -0

What is the output of the below given C program?

#include <stdio.h>
int main()
{
int loops = 0;
int _shift = 7916;
while(_shift)
{
loops++;
_shift >>= 2;
}
printf("%d",loops);
}

7-7

74-71

77-76

79-72

Solution

The binary representation of 7916 is 1111011101100

Iteration number	while	loops	Binary equivalent
1st iteration	true	1	0011110111011
2nd iteration	true	2	0000111101110
3rd iteration	true	3	0000001111011
4th iteration	true	4	0000000011110
5th iteration	true	5	0000000000111
6th iteration	true	6	0000000000001
7th iteration	true	7	0000000000000
8th iteration	false (while loop terminated)

loops variable is 7. Therefore the output is 7.

Question 8
1 / -0

Consider the C program given belwo
#include <stdio.h>
struct charNode
{
char a, b, c, d;
};
int main()
{
struct charNode ch = {'o'+3,'e'+10,'j'+15,'t'+1};
struct charNode *ptr = &ch;
printf("%c, %c",*(char *)ptr + 1,*((char *)ptr + 2));
}
What is the output?

A
'o'+ 4, 'j'+15

B
's', 'y'

C
s, y

D
'o+ 4', 'j+15'

Solution

charNode:
Data
'o'+3
'e'+10
'j'+15
't'+1
Assume address
200
201
202
203

ch points to 200
ptr = ch = 200
(char *)ptr = 200
'o' + 3 = r
*(char *)ptr + 1 = r +1 = 's'
*((char *)ptr + 2)) = *(202)
'j' + 15 = y
Output = s, y
Question 9
1 / -0

Consider the following definition of array in C:
int a [ ] [2] [3]=
      {
         {
            {1, 2, 3},
            {4, 5, 6}
         },
         {
           {10, 20, 30},
           {40, 50, 60}
        },
        {
           {100, 200, 300},
           {400, 500, 600}
   }
};
Which of the following expressions would give value 1 for the array defined above?
(a) a[0] [0] [0]
(b) * * * a
(c) * * * a[0]
(d) * * a[0] [0]
The correct answer is :

A
(a) and (c)

B
(a) and (d)

C
(a) and (b)

D
(c) and (d)

Solution

Array representation is somewhat like this:
1
2
3
10
20
30
100
200
300
4
5
6
40
50
60
400
500
600

(a) a[0] [0] [0]
When when index value is 0 , it means we are accessing the 1^st element . Here it is one. So, a[0][0][0] will give the element 1.
(b) * * * a
In case of 1 D array, single * will lead to element at a[0] position. In 2 –D array if we use “**” then it will lead us to element at a[0][0]. In similar way, ***a will give value 1 which is a[0][0][0].
(c) * * * a[0]
This is incorrect form. It will not give value 1. When there is *** a already giving value 0. No need to consider a[0] with it. It will lead to one element inside.
(d) * * a[0] [0]
It is similar to option c) which is incorrect form for output as 1. It will not result in value 1.

Question 10

1 / -0

Consider the below given code.

#include<stdio.h>

int main()

{

static int x[ ] = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10};

int i;

for(int i = 2; i < 7; i++)

x[x[i]] = x[i];

int sum = 0;

for(int j = 0; j < 10;j++)

sum = sum + x[j];

printf("%d",sum);

}

The value printed of sum is ____.

52-52

524-521

527-526

529-522

Solution

Array:

Index	0	1	2	3	4	5	6	7	8	9
Value	1	2	3	4	5	6	7	8	9	10

value of i (i<7)	operation	value modified
2	x[x[2]] = x[2]; x[3] = 3;	a[3] = 3
3	x[x[3]] = x[3] x[3] = x[3] x[3] = 3	a[3] = 3
4	x[x[4]] = x[4] x[5] =5	a[5] = 5
5	x[x[5]] = x[5] x[5] =5	a[5] = 5
6	x[x[6]] = x[6] x[7] = 7	x[7] = 7

Modified Array:

Index	0	1	2	3	4	5	6	7	8	9
Value	1	2	3	3	5	5	7	7	9	10

sum = 1 + 2 + 3 + 3 + 5 + 5 + 7 + 7 + 9 + 10 = 52.

Output: 52.

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array1	2	5	8	12	6	9	\0
Address	101	102	103	104	105	106	107
array2	-12	4	6	-13	7	\0
Address	201	202	203	204	205	206
array3	-8	-6	-7	9	\0
Address	301	302	303	304	304
array4	21	23	27	15	0	\0
Address	401	402	403	404	405	406

Variable	p	q
value stored	20	19
Assume address	1000	1002

Pointer Variable	co	vid
address stored	1002	1000

Pointer Variable	co	vid
address stored	1000	1000

Variable	p	q
Value	19	19

Programming and Data Structures Test 1

Result

Programming and Data Structures Test 1

Score

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Rank

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SHARING IS CARING

Question 1

Testbook TestSeries203

Testbook TestSeries192

Testbook TestSeries11

Error Cannot assign prinft to integer

Solution

Question 2

(a) – (iv), (b) – (ii), (c) – (iii), (d) – (i)

(a) – (i), (b) – (iv), (c) – (ii), (d) – (iii)

(a) – (ii), (b) – (iii), (c) – (i), (d) – (iv)

(a) – (iii), (b) – (i), (c) – (iv), (d) – (ii)

Solution

Question 3

GATE CS 2021

GATE CS 2022

GATE CS 2020 GATE CS 2021

It will not produce any output

Solution

Question 4

1919-1919

19194-19191

19197-19196

19199-19192

Solution

Question 5

-6 and address of 2nd element in array3

-5 and 23

-6 and 23

-5 and address of 2nd element in array4

Solution

Question 6

(a) and (b)

(a) and (d)

(b) and (c)

(c) and (d)

Solution

Question 7

7-7

74-71

77-76

79-72

Solution

Question 8

'o'+ 4, 'j'+15

's', 'y'

s, y

'o+ 4', 'j+15'

Solution

Question 9

(a) and (c)

(a) and (d)

(a) and (b)

(c) and (d)

Solution

Question 10

52-52

524-521

527-526

529-522

Solution

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Testbook TestSeries
20
3

Testbook TestSeries
19
2

Testbook TestSeries
1
1

-6 and address of 2^nd element in array3

-5 and address of 2^nd element in array4