GATE CSE 2016 Set 1 | Question: 35

Question

 What will be the output of the following $C$ program?

void count (int n) {
    static int d=1;
    
    printf ("%d",n);
    printf ("%d",d);
    d++;
    if (n>1) count (n-1);
    printf ("%d",d);
    
    
}

void main(){
    count (3);
}

• A. $3 \ 1 \ 2 \ 2 \ 1 \ 3  \ 4 \ 4 \ 4$ 
• B.  $3 \ 1 \ 2  \ 1  \ 1 \ 1 \ 2 \ 2 \ 2$ 
• C.  $3 \ 1  \ 2 \ 2 \ 1 \ 3 \ 4$ 
• D.  $3 \ 1 \ 2  \ 1  \ 1  \ 1  \ 2$

Comments

No, I think that is not the else part. It is the statement which will execute irrespective of the above if condition.

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Isn't  printf("%d",d);  the ELSE part ?

if (n>1) count (n-1);   <-  If condition true 
    printf ("%d",d);    <- If condition false

Then C will be correct answer.

how can we know that ???,,,,,while practicing i also thought it coulb be else part

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If you look carefully , the program has 3 print statements without any condition.

So the outputs generated will be multiple of 3.

$\therefore$ Option $c$ and $d$ are eliminated.(as they generate 7 outputs)

Answer 1

Guys reccursion tree:

https://gateoverflow.in/?qa=blob&qa_blobid=11408070808494714478

Answer 2

D is the static variable so the answer is an option (A) 3 1 2 2 1 3 4 4 4

1. count(3): Print 3 1

     d=2,count(2)

2. count(2): print 2 2

     d=3,count(1)

3. count(1): print 3 1

Now the condition fails and it will print value 4, for 3 times.

Answer 3

Answer:  option (a) =>  312213444

first two prints for count (3) call
n=3 d=1

first two prints for count (2) call 
n=2 d=2 

first two prints for count (1) call
n=1 d=3

After that it will follow recursion stack to print remaining print statements.

Since 'd' is a static variable so value is shared among the recursive calls and it will print the latest updated value of d that is 3 times 4.. d=4 after evaluation of  d++ in last recursive call.