GATE CSE 1998 | Question: 2.12

Question

What value would the following function return for the input $x=95$?

Function fun (x:integer):integer;
Begin
    If x > 100 then fun = x – 10
    Else fun = fun(fun (x+11))
End;

• A. $89$
• B. $90$
• C. $91$
• D. $92$

Comments

How I can do this? I mean which offline ide you are using and how to check the call stack after execution?

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He is using code blocks. You can do it in any IDE. Most of them have a debugging feature where you can check the call stack.

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starting point is fun(95)

Answer 1

Value returned by $\text{fun}(95) = \text{fun}(\text{fun}(106))$
$\qquad\qquad= \text{fun}(96)$
$\qquad\qquad= \text{fun}(\text{fun}(107))$
$\qquad\qquad= \text{fun}(97)$
$\qquad\qquad= \text{fun}(\text{fun}(108))$
$\qquad\qquad= \text{fun}(98)$
$\qquad\qquad = \text{fun}(\text{fun}(109))$
$\qquad\qquad= \text{fun}(99)$
$\qquad\qquad= \text{fun}(\text{fun}(110))$
$\qquad\qquad= \text{fun}(100) $
$\qquad\qquad= \text{fun}(\text{fun}(111)) $
$\qquad\qquad= \text{fun}(101) = 91.$

Correct Answer: $C$

Comments

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

#include <iostream> using namespace std; int ans; int fun(int no){ if(no > 100){ ans = no - 10; // this is the terminating condition return no-10; } else{ return fun(fun(no+11)); } } int main() { fun(95); cout << ans << "\n"; return 0; }

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@neel19 how after 91 the fucntion is not calling fun(102) ?? can u plz elaborate it 

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If you see the 3^rd last step, we have fun(100). The code says :- 

If x > 100 then fun = x – 10
    Else fun = fun(fun (x+11))

 

Now fun(100) = fun(fun(111)) // Since 100 is not > 100.

= fun(101) // since 111 > 100

 now this is unlike the other terms which were < 100, so we need to add 11. 

Here 101 > 100, thus we again subtract 10 from 101

→ fun = 101-10 = $91$

Answer 2

Yes 91 is correct.

step 1: when fun=111 -> fun(fun(111))

step 2:  fun=101 ->fun(101)

step 3:As 101>100 fun=fun -10 =101-10=91

step 4: as there is no fun function call remaining ,it will exit.

Comments

how can we understand this is stop codition.

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Because when value is 111, there are no further function call. Just the return.

Answer 3

It might help.

int fun(int x)
{
     if(x>100)
     return (x-10);
     else
     return fun(fun(x+11));
}

Equivalent function.

Answer 4

I think the answer is 91.

Comments

Thanks Ahsanul