GATE CSE 2017 | Set 1 | Question: 35

Question

Consider the following two functions.

void fun1(int n) {
    if(n == 0) return;
    printf("%d", n);
    fun2(n - 2);
    printf("%d", n);
}
void fun2(int n) {
    if(n == 0) return;
    printf("%d", n);
    fun1(++n);
    printf("%d", n);
}

The output printed when $\text{fun1}(5)$ is called is

• A. $53423122233445$  
• B. $53423120112233$                       
• C. $53423122132435$  
• D. $53423120213243$                       

Comments

The key point here is f1(n-2) is only modifying the value so it just print n where f2(n++) is incrementing after the call so it will be printing n+1 during the return step

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THIS EXAMPLE SHOWS one should always look options and then solve the question
if u did the same the same u will save alot of time
just after 2 steps u got ur answer

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https://www.youtube.com/watch?v=jWU9sfLDDIo
detailed explanation for the question

Answer 1

• Unroll recursion up to a point where we can distinguish the given options and choose the correct one!
• Options B and D are eliminated.
• A is the answer.

Comments

pre-increment and for every function call there’ll be activation record which will store the local variable values .So, at the time of fun call ,variable value has been incremented .It was stored and that’s the reason why it has been incremented.

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Nice Approach Diddy

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@Aashish

because it is pre incremented in n . and that n is passed to pun1(n),so when we retrn back ,in its stak n is equivalent to 4

Answer 2

hence correct answer is a.

Comments

Thank you :)

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please anyone can tell me that why right side printf value is incremented

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Ray Tomlinson in $f_2()$ after the printf statement there is a pre-increment of variable $n$.

Answer 3

ANS)A

In fun2,value of n also get incremented after each function call.

Answer 4

\

Solution......

Comments

please anyone can tell me that why right side printf value is incremented

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because of ++n inside the  fun2() : pre increment

Answer 5

Ans is A 

Comments

I have another query...fun2(++n) and fun2(n++)----------- Both are same right?

Here, we won't check pre/post increment?

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in fun2(++n) , let take fun2(3) , value of n=3 and its ++n so first update the value of n , become n=4 and then passed as fun1(4)

but in fun2(n++) , let take same example fun2(3) , value of n=3 and its n++ so first passed value as fun1(3) then change value of n=4

( here n is local variable )

Answer 6

A is the correct answer

Comments

I think for weak students like me step by step just follow the number

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why we have printed 2 in the leaf node right hand side instead of 0?

 

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now see u will understand