Programming GATE 2022 Solved Questions

Ques 1: What is printed by the following ANSI C program?

GATE 2022 Q no 21

#include<stdio.h>
int main(int argc, char *argv[])
{
 int x = 1, z[2] = {10, 11};
 int *p = NULL;
 p = &x;
 *p = 10;
 p = &z[1];
 *(&z[0] + 1) += 3;
 printf("%d, %d, %d\n", x, z[0], z[1]);
 return 0;
}

(A) 1 ,10 ,11

(B) 1 ,10 ,14

(C) 10 ,14 ,11

(D) 10 ,10 ,14

Ans: (D) 10 ,10 ,14

Solution: Initially p is holding the address of x

So, *p=10 will change the value of x to 10

After that p will hold the address of z[1]

*(&z[0] + 1) += 3 will change the contents of z[0] +1 i.e z[1] incremented by 3

But contents of z[0] remains unchanged

Finally, output will be 10, 10, 14

Ques 2: What is printed by the following ANSI C program?

GATE 2022 Q no 43

#include<stdio.h>
int main(int argc, char *argv[])
{
 int a[3][3][3] =
 {{1, 2, 3, 4, 5, 6, 7, 8, 9},
 {10, 11, 12, 13, 14, 15, 16, 17, 18},
 {19, 20, 21, 22, 23, 24, 25, 26, 27}};
 int i = 0, j = 0, k = 0;
 for( i = 0; i < 3; i++ ){
 for(k = 0; k < 3; k++ )
 printf("%d ", a[i][j][k]);
 printf("\n");
 }
 return 0;
}

Ans: (A)
1 2 3
10 11 12
19 20 21

Solution: a is 3D array with size [3][3][3]

Therefore each 2D array contains 9 elements, we have 3 such arrays

0th 2D array have {1,2,3,4,5,6,7,8,9}

1st 2D array have {10,11,12,13,14,15,16,17,18}

2nd 2D array have {19,20,21,22,23,24,25,26,27}

each 2D array is collection of 1D array.

We have 3 one dimensional arrays in one 2D

when i=0, j=0, output = {1,2,3}

when i=1, j=0, output = {10,11,12}

when i=2, j=0, output = {19,20,21}

Ques 3: What is printed by the following ANSI C program?

GATE 2022 Q no 44

#include<stdio.h>
int main(int argc, char *argv[]){
 char a = 'P';
 char b = 'x';
 char c = (a & b) + '*';
 char d = (a | b) - '-';
 char e = (a ^ b) + '+';
 printf("%c %c %c\n", c, d, e);
 return 0;
}

(A) z K S

(B) 122 75 83

(C )∗ – +

(D) P x +

Ans: (A) z K S

Solution: char a = ‘P’

char b = ‘x’

As per precedence of operators, () evaluated prior to +/-

Note that &,^ and | are bit wise operators and Addition/Subtraction of characters applied on their ASCII values.

a = ‘P’ means a = ASCII value of (‘P’) = 65+15 = 80 = (0101 0000)₂

b = ‘x’ means b = ASCII value of (‘x’) = 97+23 = 120 = (0111 1000)₂

a & b    = (0101 0000)₂  [ apply & logic on corresponding bits ] = (80)₁₀

‘*’ = 42 = (0010 1010)₂

a&b + ‘*’ = 80 + 42 = 122

print character ( 122 ) = z

a | b    = (0111 1000)₂  [ apply | logic on corresponding bits ] = (120)₁₀

‘-’ = 45

a|b – ‘-’ = 120 – 45 = 75

print character ( 75 ) = K

a ^ b    = (0010 1000)₂  [ apply ^ logic on corresponding bits ] = (40)₁₀

‘+’ = 43

a^b + ‘+’ = 40 + 43 = 83

print character ( 83 ) = S