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

In Go book ,this questions  are putted in algorithms  ,should  be under recursion  in c programming

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.

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 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

Answer 3

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