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

Whenever we take a static variable, until function calls are over we won't print the value i.e. we take the last updated value after the function call and then we print it,but if the variable is global then why do we directly take the value and print it .​​?Here second printf stmt i.e.stmt after function call we fixed the value into the variable and it's been printed at the end.why? 

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https://www.youtube.com/watch?v=V7n6aFiBb7s

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Yeah got it bro.And the video is primarily focussed on how to use options to get the correct solution

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Mistakes we might do in the question are like not considering pre-increment.

Answer 1

Question can be answered in 20-30 sec

Answering to this question is very simple(elimination technique).

fun1(){

print(%d)// prints value

fun2(n-2)

print(%d)// prints same value

}

if we analyze this properly value wont change in function calling of fun1{ fun2(n-2) }.

 

first and second print statement will print same value

so starting value is 5 and ending value will be even 5.

Therefore b & d are eliminated.

fun2(){

print(%d)// prints value

fun1(++n)

print(%d)// prints Incremented value because of pre increment

}

so by analyzing this we can eliminate c.

Answer is A

Answer 2

For fun1(5)

fun1(5)➣ fun2(3) ➣fun1(4) ➣fun2(2)➣ fun1(3) ➣fun2(1) ➣fun1(2) ➣fun2(0)

   ↓               ↓              ↓             ↓              ↓             ↓               ↓            ↓

   5               3              4              2                3             1                2            2 

                                                                                                                    [return to fun2(n-2)   printf("%d",n) ➣ n-2=0 ➣ n=2] 

  ↓                    ↓              ↓                       ↓                 ↓                 ↓

  5                   4              4                      3                  3                  2        

 now return values to fun1 , fun2       

for fun1 return n   

for fun2  return n+2 Bcz   fun2(n-2) instruction pointer on 'n' that's n   .

so final answer is : option A : 53423122233445 

                                                           

 

Answer 3

Here, $\implies$ means print.

fun1(0) $\implies$ {nothing}

fun2(0) $\implies$ {nothing}

fun1(2) $\implies$ 2 {fun2(0)} 2 $\implies$ $22$

fun2(1) $\implies$ 1 {fun1(2)} 2 $\implies$ $1222$

fun1(3) $\implies$ 3 {fun2(1)} 3 $\implies$ $312223$

fun2(2) $\implies$ 2 {fun1(3)} 3 $\implies$ $23122233$

fun1(4) $\implies$ 4 {fun2(2)} 4 $\implies$ $4231222334$

fun2(3) $\implies$ 3 {fun1(4)} 4 $\implies$ $342312223344$

fun1(5) $\implies$ 5 {fun2(3)} 5 $\implies$ $53423122233445$

Answer :- A.