Copy Of Quiz 2 For Section L

1. What is the output of this C code?

  #include <stdio.h>
    int main()
    {
        do
            printf("In while loop ");
        while (0);
            printf("After loop\n");
    }

In while loop

In while loop After loop

After loop

For (i = n-1; i>-1; i–)

The code uses a do-while loop with a condition of 0, which means the loop will execute at least once. Inside the loop, it prints "In while loop ". After the loop, it prints "After loop". Therefore, the output of the code will be "In while loop After loop".

Explanation

The code uses a do-while loop with a condition of 0, which means the loop will execute at least once. Inside the loop, it prints "In while loop ". After the loop, it prints "After loop". Therefore, the output of the code will be "In while loop After loop".

2. What is the output of this C code? #include int main() { int x = 1; if (x > 0) printf("inside if\n"); else if (x > 0) printf("inside elseif\n"); }

Inside if

Inside elseif

Inside if inside elseif

Syntax error

The output of this C code is "inside if". This is because the variable x is equal to 1, which is greater than 0. Therefore, the condition in the if statement is true and the code inside the if block is executed. The else if statement is not evaluated because the condition x > 0 is already satisfied.

Explanation

The output of this C code is "inside if". This is because the variable x is equal to 1, which is greater than 0. Therefore, the condition in the if statement is true and the code inside the if block is executed. The else if statement is not evaluated because the condition x > 0 is already satisfied.

3.

What is the output of the given code:

#include <stdio.h>
  void main()
  {

       int i = 0;

       while (++i)
 {
  printf("H");
     }
 }

H

Error

H is printed infinite times

None

The code contains an infinite loop. The condition in the while loop is "++i", which means that the value of "i" will be incremented by 1 before each iteration of the loop. Since the condition is always true (non-zero), the loop will continue indefinitely. Inside the loop, the code prints "H" repeatedly, resulting in "H" being printed infinite times.

Explanation

The code contains an infinite loop. The condition in the while loop is "++i", which means that the value of "i" will be incremented by 1 before each iteration of the loop. Since the condition is always true (non-zero), the loop will continue indefinitely. Inside the loop, the code prints "H" repeatedly, resulting in "H" being printed infinite times.

4. How many times is a do while loop guaranteed to loop?

1

0

Infinite

Depend on Variable

A do while loop is guaranteed to loop at least once because the condition is checked at the end of the loop. This means that the loop will always execute the code block once before checking if the condition is true or false. Therefore, even if the condition is false from the beginning, the loop will still run once.

Explanation

A do while loop is guaranteed to loop at least once because the condition is checked at the end of the loop. This means that the loop will always execute the code block once before checking if the condition is true or false. Therefore, even if the condition is false from the beginning, the loop will still run once.

5. What is the output of this C code? #include <stdio.h> int main() { int i = 4, j = 7; do { printf("Loop"); }while(++i < --j); return 0; }

Loop

LoopLoop

Prints nothing

Compilation Error

The code uses a do-while loop to repeatedly print "Loop" as long as the condition (++i

Explanation

The code uses a do-while loop to repeatedly print "Loop" as long as the condition (++i

6.

What is the output of the given code:
int main()
{
    int i;
    int arr[5] = {1};
    for (i = 0; i < 5; i++)
        printf("%d ", arr[i]);
    return 0;
}

1 followed by four garbage values:

10000

11111

00000

The given code initializes an integer array `arr` with size 5 and assigns the first element as 1 while leaving the rest of the elements uninitialized. The code then enters a for loop that iterates 5 times. Each iteration, it prints the value of `arr[i]`, which in this case will be 1 for the first iteration and garbage values for the rest of the iterations since the rest of the elements in the array are uninitialized. Therefore, the output will be "1 followed by four garbage values".

Explanation

The given code initializes an integer array `arr` with size 5 and assigns the first element as 1 while leaving the rest of the elements uninitialized. The code then enters a for loop that iterates 5 times. Each iteration, it prints the value of `arr[i]`, which in this case will be 1 for the first iteration and garbage values for the rest of the iterations since the rest of the elements in the array are uninitialized. Therefore, the output will be "1 followed by four garbage values".

7. What is the output of this C code? #include <stdio.h> int main() { int n; for (n = 9; n!=0; n--) printf("n = %d", n--); return 0; }

9 7 5 3 1

9 8 7 6 5 4 3 2 1

Infinite

Error

The code starts with n = 9 and enters a for loop. In each iteration of the loop, it prints the value of n and then decrements n by 2. However, the decrement operator (--) is used twice in each iteration, once in the printf statement and once in the loop itself. This leads to n being decremented by 2 in each iteration instead of 1. As a result, n will never be equal to 0, causing an infinite loop. Therefore, the output of the code is infinite.

Explanation

The code starts with n = 9 and enters a for loop. In each iteration of the loop, it prints the value of n and then decrements n by 2. However, the decrement operator (--) is used twice in each iteration, once in the printf statement and once in the loop itself. This leads to n being decremented by 2 in each iteration instead of 1. As a result, n will never be equal to 0, causing an infinite loop. Therefore, the output of the code is infinite.

8.

What is the output of this C code?
#include <stdio.h> 
 int main()
{
 int x = 0;
  if (x++)
 printf("true\n");

       else if (x == 1)

            printf("false\n");

True

False

Truefalse

0 time

The output of this C code is "false". This is because the initial value of x is 0. In the if statement, x is incremented using the post-increment operator, but the value of x before the increment is evaluated in the condition. Since the value of x before the increment is 0, the condition evaluates to false and the corresponding printf statement is executed.

Explanation

The output of this C code is "false". This is because the initial value of x is 0. In the if statement, x is incremented using the post-increment operator, but the value of x before the increment is evaluated in the condition. Since the value of x before the increment is 0, the condition evaluates to false and the corresponding printf statement is executed.

9. What is the output of this C code? int main () { int i=0; for (i=0; i<20; i++) { switch (i) { case 0 : i += 5; case 1 : i += 2; case 5 : i += 5; default : i += 4; break; } printf("%d, ", i); } }

0,5,9,13,17

5,9,13,17

12,17,22

16,21

The code starts with initializing the variable i to 0. Then, it enters a for loop that iterates until i is less than 20. Inside the loop, there is a switch statement that checks the value of i.

When i is 0, it adds 5 to i and continues to the next case. When i is 1, it adds 2 to i and continues to the next case. When i is 5, it adds 5 to i and continues to the next case.

In the default case, it adds 4 to i and then breaks out of the switch statement. Finally, the value of i is printed using printf.

The output of the code is 16,21 because after the loop iterates 16 times, the value of i becomes 16 and then it adds 4 in the default case before printing it.

Explanation

The code starts with initializing the variable i to 0. Then, it enters a for loop that iterates until i is less than 20. Inside the loop, there is a switch statement that checks the value of i.

When i is 0, it adds 5 to i and continues to the next case. When i is 1, it adds 2 to i and continues to the next case. When i is 5, it adds 5 to i and continues to the next case.

In the default case, it adds 4 to i and then breaks out of the switch statement. Finally, the value of i is printed using printf.

The output of the code is 16,21 because after the loop iterates 16 times, the value of i becomes 16 and then it adds 4 in the default case before printing it.

10. What is the output of this piece of code? #include <stdio.h> void main() { int i; for(i=0;i<10;i++) printf("%d ",i); }

0 1 2 3 4 5 6 7 8 9

0 1 2 3 4 5 6 7 8 9 10

10

Error

The given code is a simple for loop that iterates from 0 to 9. Inside the loop, it prints the value of 'i' followed by a space. After the loop completes, it prints the value of 'i' one more time, which is 10. Therefore, the output of the code is "0 1 2 3 4 5 6 7 8 9 10".

Explanation

The given code is a simple for loop that iterates from 0 to 9. Inside the loop, it prints the value of 'i' followed by a space. After the loop completes, it prints the value of 'i' one more time, which is 10. Therefore, the output of the code is "0 1 2 3 4 5 6 7 8 9 10".