## Control Flow

 Question 1
What is the following program segment doing?
main()
{
int d=1;
do
{
printf(“%d”\n”,d++);
}while(d<=9);
}
 A Adding 9 integers B Adding integers from 1 to 9 C Displaying integers from 1 to 9 D No output
Question 1 Explanation:
The code consists of do-while loop in which action performs first and later condition checking.
In the printf() statement, d++ means first it will display d value and increment the d value later condition checking.So the integer values 1 to 9 will be printed.
 Question 2
How many lines of output does the following C code produce?
#include<stdio.h>
float i=2.0;
float j=1.0;
float sum = 0.0;
main()
{
while (i/j > 0.001)
{
j+=j;
sum=sum+(i/j);
printf("%f\n", sum);
}
}
 A 8 B 9 C 10 D 11
Question 2 Explanation:
Iteration-1:
while (1.000000 > 0.001)
{
j=2.0
sum=0+1.000000;
printf("%f\n",sum); /* It will print 1.000000 */
}
Iteration-2: 1.500000
Iteration-3: 1.750000
Iteration-4: 1.875000
Iteration-5: 1.937500
Iteration-6: 1.968750
Iteration-7: 1.984375
Iteration-8: 1.992188
Iteration-9: 1.996094
Iteration-10: 1.998047
Iteration-11: 1.999023
The program will terminate after 11th iteration. So, it print 11 lines.
 Question 3
Consider the following segment of C-code:

The number of comparisons made in the execution of the loop for any n > 0 is:
 A ⌊log 2n⌋*n B n C ⌊log 2n⌋ D ⌊log 2n⌋+1
Question 3 Explanation:
Explanation:
Let us consider n=6, then
1<=6 (correct)
2<=6 (correct)
4<=6 (correct)
8<=6 (False)
4 comparisons required
Option A:
⌊log n⌋+1
⌊log 6⌋+1
3+1=4 (correct)
Option B:
n=6 (False)
Option C:
⌊log n⌋
⌊log 6⌋=3 (False)
Option D:
⌊log 2n⌋+1
⌊log 26⌋+1=2+1=3 (False)
 Question 4

The complexity of the program is
 A O(log n) B O(n2) C O(n2 log n) D O(n log n)
Question 4 Explanation:
 Question 5
What will be the output of the following C code?
 A 10 11 12 13 14 B 10 10 10 10 10 C 0 1 2 3 4 D Compilation error
Question 5 Explanation:
Step-1: We are initialized i=0 in for loop. It means condition true because it is less than 5.
Step-2: Inside the for loop we are assigning again I value is 10 and printing value i.
Iteration-1: We are printing value 10 then increment by 1
Iteration-2: We are clearing previous value and assigning value 10. Printing value is 10;
Iteration-5: We are clearing previous value and assigning value 10. Printing value is 10
Step-3: output is 1010101010
There are 5 questions to complete.

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