What will be the output of the following code snippet?

#include <stdio.h>

void solve() {

int a[] = {1, 2, 3, 4, 5};

int sum = 0;

for(int i = 0; i < 5; i++) {

if(i % 2 == 0) {

sum += *(a + i);

}

else {

sum -= *(a + i);

}

}

printf("%d", sum);

}

int main() {

solve();

return 0;

}

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• Syntax error

• 3

What is the disadvantage of arrays in C?

• The amount of memory to be allocated should be known beforehand.

• Elements of an array can be accessed in constant time.

• Elements are stored in contiguous memory blocks.

• Multiple other data structures can be implemented using arrays.

What will be the output of the following code snippet?

#include <stdio.h>

void solve() {

int a = 3;

int res = a++ + ++a + a++ + ++a;

printf("%d", res);

}

int main() {

solve();

return 0;

}

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• 20

• 18

If p is an integer pointer with a value 1000, then what will the value of p + 5 be?

• 1020

• 1005

• 1004

• 1010

What will be the result of the following code snippet?

#include <stdio.h>

void solve() {

char ch[10] = "abcdefghij";

int ans = 0;

for(int i = 0; i < 10; i++) {

ans += (ch[i] - 'a');

}

printf("%d", ans);

}

int main() {

solve();

return 0;

}

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• 20

• 100

Which of the following is not a storage class specifier in C?

• Volatile

• Extern

• typedef

• Static

What kind of linked list is best to answer questions like “What is the item at position n?”

• Singly linked list

• Doubly linked list

• Circular linked list

• Array implementation of linked list

Which of the following does the balancing symbols algorithm include?

• balancing double quotes

• balancing single quotes

• balancing operators and brackets

• balancing parentheses, brackets and braces

Consider the graph shown below.

Which of the following edges form the MST of the given graph using Prim’a algorithm, starting from vertex 4.

• (4-3)(5-3)(2-3)(1-2)

• (4-3)(3-5)(5-1)(1-2)

• (4-3)(3-5)(5-2)(1-5)

• (4-3)(3-2)(2-1)(1-5)

What will be the output of the following code?

#include<bits/stdc++.h>

using namespace std;

void func(char* str2, char* str1)

{

int m = strlen(str2);

int n = strlen(str1);

for (int i = 0; i <= n - m; i++)

{

int j;

for (j = 0; j < m; j++)

if (str1[i + j] != str2[j])

break;

if (j == m)

cout << i << endl;

}

}

int main()

{

char str1[] = "1253234";

char str2[] = "323";

func(str2, str1);

return 0;

}

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