 # explain output

what will be the output of following code:

``````void fun(int x)
{
if(x>0)
{
fun(--x);
printf("%d ",x);
fun(--x);
}
}
int main()
{
int a=4;
fun(a);
getch();
return 0;
``````

}

Before explaining you the output for

``````fun(4)
``````

, let us see outputs for numbers less than 4 -:
Let us denote the ouput for

``````fun(n)
``````

as " out(n) "

**1).

``````fun(0)
``````

**

In this case ,there will be no output as the

``````if(n>0)
``````

condition is not satisfied…
Therefore, out(0) = ( Nothing/Blank output )

**2).

``````fun(1)
``````

**

In this case, the condition is satisfied, so

``````fun(--x)
``````

will be executed.
Firstly the value of x will decrease by 1, making x=0.
Now

``````fun(0)
``````

``````fun(0)
``````

's output and denote it as " out(0) " and continue executing the function…
The next line is

``````printf("%d ",x);
``````

=> “0” will be the output of this printf statement…
And now here comes our second

``````fun(--x)
``````

statement.
Now, first x will become = -1 ( It had become 0 earlier… ) and

``````fun(-1)
``````

will be called. ( whose output , just like

``````fun(0)
``````

, will again be nothing)
So the final output would be-:
= out(1)
= out(0) + 0 + out(-1)
= nothing + 0 + nothing
=> 0

**3).

``````fun(2)
``````

**

In the same way as mentioned earlier the first output would be “out(1)” ( x became 1 from 2 before it got passed as an argument for

``````fun(--x)
``````

), …
Then the printf statement would output “1” …
And then the third output would be “out(0)” ( Again, x became 0 from 1 before being passed as an argument)

Therefore the final output -:
= out(2)
= out(1) + 1 +out(0)
= 0 + 1 + nothing
=> 0 1

**4).

``````fun(3)
``````

**

Similar to the previous case, its output would be -:

= out(3)
= out(2) + 2 + out(1)
= 0 1 + 2 + 0
=> 0 1 2 0

**5).

``````fun(4)
``````

**

Following the same analogy ,

= out(4)
= out(3) + 3 + out(2)
= 0 1 2 0 + 3 + 0 1
=> 0 1 2 0 3 0 1

//