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;
}
please explain it too!
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)
will be called…Now for a moment, let us forget about this
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