GATE CSE 2024 | Set 1 | Question: 8

Question

Consider the following $\mathrm{C}$ program:

#include <stdio.h>

int main() {

int a=6;

int b = 0;

while (a<10)   {

a = a / 12+1 ;

a += b ;}

printf ("%d", a);

return 0 ; }

Which one of the following statements is CORRECT?

• A. The program prints $9$ as output
• B. The program prints $10$ as output
• C. The program gets stuck in an infinite loop
• D. The program prints $6$ as output

Comments

In first iteration 6<10 then in the last a becomes 1 after that it will check for 1<10 thus a becomes 1 and after that it stucks in an infinite loop for every iteration value of a becomes 1 so answer is

Option C

Answer 1

The initial value is $a=6,b=0$. When the first time while loop is runs the conditions become true as $(6<10)$. Now control enters inside the for loops and executes the following lines:

$a=a/12+1\implies (6/12)+1=0+1=1$: it returns $a=1$

$a=a+b \implies1+0=1$: it returns again $a=1$

Now in the second iteration again while conditions become true as $(1<10)$ and again control enters inside the loops and executes the inside statements:

$a=a/12+1$: it returns $a=1$

$a=a+b$: it returns again $a=1$

If we execute again and again every time conditions become true. So no output is printed and programs go into infinite loops.

Option $(C)$ is correct.

Comments

We can notice: a+=b is dead code, i.e., no use with that as b =0 initially and not changing in the loop.

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where it returns A,it  returns always zero

Answer 2

Inside the while loop a is a dependent variable that while loop will use for it's recurrence.

and we have a as,

a = a/12 + 1;

a += b; 

with a = 6 and b = 0 as initial values.

so, a = 6/12 + 1 = 0.5 + 1 = 0 + 1 [because a is an integer variable where any values after decimal point are simply dropped.]

and then a = 1 + 0 = 1

similarly more runs of a will make the value of a as 1 only. Because of the integer division.

Hence the program will never end. Option C.

Answer 3

In 1st iteration,

a = 6/10 +1
a += 0
Since a is an integer, a = 1 at the end of first iteration

In 2nd iteration,

a = 1/10 +1
a += 0
Since a is an integer, a = 1 at the end of second iteration

So, value of a will always be less than 10, the while loop will never end.

Answer: option C

Answer 4

answer is Option(C)

 

given a = 6

b = 0

condition is while ( a<10)

 

a = a/12+1;

a = 6/12+1 = 0.5 +1 = 0 +1 = 1 (because its integer variable s we take 0 not 0.5)

now a += b :   a = a+b = 1+0 = now value of a = 1;

now, value of a =1  which will be always be less than 10, the while loop will never ends

 

Answer 5

c. The program gets stuck in an infinite loop

Answer 6

here if we observe closely we find there is no increment in veriable a

so that we can not fail the loop condition thats why it run infinitely

note here there is no chance of stack over flow because there is no function call

when function call happen like this the after certain point of time stack exhusted