GATE IT 2008 | Question: 83

Question

Consider the code fragment written in C below :
        

void f (int n)
{ 
    if (n <= 1)  {
        printf ("%d", n);
    }
    else {
        f (n/2);
        printf ("%d", n%2);
    }
}

Which of the following implementations will produce the same output for $f(173)$ as the above code?

P1	P2
void f (int n) { if (n/2) { f(n/2); } printf ("%d", n%2); }	void f (int n) { if (n <=1) { printf ("%d", n); } else { printf ("%d", n%2); f (n/2); } }

• A. Both $P1$ and $P2$
• B. $P2$ only
• C. $P1$ only
• D. Neither $P1$ nor $P2$

Comments

void fun() {
static int n=173;
if(n==0) return;

n=n/2;

fun(n);
printf("%d" , n%2);

} actually i was thinking abt this case static is there it will always Print 0

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Static int n; --- this line demands a compile time constant. It can't take the value at the run time.

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P2 is printing in reverse binary represntation

Answer 1

Answer: C

The code fragment written in C and P1 prints the binary equivalent of the number n.

P2 prints the binary equivalent of the number n in reverse.

Comments

Isn't p1 is also printing one extra 0 at starting and the representation becomes 010101101

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Let's take 5 as input. Then P1 printing that

N= 5 ---- this will call f(2) and append 1 at the end

N=2 --- this will call f(1) and append 0 at the end

N=1 --- this will not enter into the if(), and directly return with 1.

 

In sum up, after returning 1, append 0, then append 1.

N=5, output = 101

Answer 2

Here, $P1$ and $P2$ will print opposite in direction as shown in diagram.

And given code fragment will print like $P1$ and not like $P2$

Hence, answer will be (C).

Comments

nice answer.

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tnks :)

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P1 is printing a leading 0,so shouldn't it be option (D) ?

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There is no leading 0 printed. On reaching f(1) in P1, there's a check if(n/2){..}. The check fails and does not result in 0 being printed since f(0) is not invoked.

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once we know what the code is doing take smaller inputs on which output is  not a palindrome so that you are able to make a difference which function is doing what

like binary of 6 is 110(This binary output is not a palindrome), but P2 prints 011, so it is wrong.

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nice approach

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Can some explain how P1 is giving output?

since if(n/2) condition first check for 173 the 86 which will be false for 86 and 0 will printed first .

correct me please.

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it is if(n/2) not if(n%2)

if(86/2) will be if(43) which is true

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@Arjun sir this is best answer than selected one!

Answer 3

ans is 3). As P1 prints 10101101 same as given code in question but P2 prints the output in reverse order.