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

Here, d is Static, so the value of d will persists between the function calls.

1.  $\text{count(3)}$ will print the value of $n$ and $d$ and increments $d$ and call $\text{count(2)} \Rightarrow \text{prints} \ 3 \ 1$.
2.  $\text{count(2)}$ will print the value of $n$ and $d$ and increments $d$ and call $\text{count(1)} \Rightarrow \text{prints} \ 2 \ 2$.
3.  $\text{count(1)}$ will print the value of $n$ and $d$ and increments $d  \Rightarrow \ \text{prints} \ 1 \ 3$.

Now, it will return and prints the final incremented value of $d$ which is $4$, three times.
So, option (A) is correct $= 3 \ 1 \ 2 \ 2 \ 1 \ 3 \ 4 \ 4 \ 4$

Comments

 srestha if d is not initialized as static then any changes in output???

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yes , then d++; will get no meaning. right??

Because in every call $d$ initialized with $1.$

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If d is not static. Then answer will be option B) 312111222.

Answer 2

A static local variable is different from a local variable as a static local variable is initialized only once no matter how many times the function in which it resides is called and its value is retained and accessible through many calls to the function in which it is declared.

If a variable is declared static in a function, the same variable/same copy will be used for all recursive calls of that function.

Static variables are allocated memory in the data segment, not stack segment. In this case, d is initialized explicitly so it will be stored in initialized data segment.

Check this - http://www.geeksforgeeks.org/memory-layout-of-c-program/

here count(3) will do 3 things  - prints 31, increments d and call count(2)

count(2) - print 22, increments d and call count(1)

count(1) -  print 13, increments d. Now as 0 $\ngtr 1$ , of it will print final value of d i.e 4 and returns to count(2). count(2) will print 4 and returns to count(3), At last count(3) will print 4.

Final output - 312213444

A is the correct answer. 

Answer 3

(A)  3 1 2 2 1 3 4 4 4

Comments

pls compile and then check.its getting A. i too did wrong. i marked c in exam.

Answer 4

Answer is A.

When the recursion ends. 4 will be printed 3 times.

Here is the running code of above question.

https://ideone.com/JDW1Py