fun(int x) { if(x > 0) fun(x/2); printf("%d", x); } ab

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fun(int x)
{
if(x > 0)
fun(x/2);
printf("%d", x);
}

above function is called as:
fun(10);

what will it print?

}

Question Posted / sid08

17 Answers
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fun(int x) { if(x > 0) fun(x/2); printf("%d", x); } above function is call..

Answer / shilpa m

Right answer is 0.
fun(int x)
{
if(x > 0)
fun(x/2);
printf("%d\n", x);
}
Here if fun(10)is called the sequence goes as fun(10)->fun
(5)->fun(2.5)->fun(1.25)->fun(0.625) after this itself
printf will be executed and 0 is printed.

please expalin how answer is 0 1 2 5 10 is right answer???

Is This Answer Correct ?

1 Yes

1 No

fun(int x) { if(x > 0) fun(x/2); printf("%d", x); } above function is call..

Answer / enigma

Strictly speaking the answer is undefined until someone forces the output text from the console buffer to the screen.
Otherwise it would normally print 012510

Debug it inside MSVC and witness no output. In reality though most implementations will do a final flush to screen....

Is This Answer Correct ?

0 Yes

0 No

fun(int x) { if(x > 0) fun(x/2); printf("%d", x); } above function is call..

Answer / mortal

it will print "0" i.e zero since compiler wont get to the
print statement until the value is zero.

Is This Answer Correct ?

2 Yes

5 No

fun(int x) { if(x > 0) fun(x/2); printf("%d", x); } above function is call..

Answer / vinay

10,5,2,1

after that it terminated.

Is This Answer Correct ?

1 Yes

4 No

fun(int x) { if(x > 0) fun(x/2); printf("%d", x); } above function is call..

Answer / sri

Is This Answer Correct ?

1 Yes

6 No

fun(int x) { if(x > 0) fun(x/2); printf("%d", x); } above function is call..

Answer / divya

Is This Answer Correct ?

1 Yes

9 No

fun(int x) { if(x > 0) fun(x/2); printf("%d", x); } above function is call..

Answer / megha

Is This Answer Correct ?

1 Yes

15 No

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