What's it's ouput? Please explain briefly.

Question

void func(int x)
{
if(x>0)
{
func (--x);
cout<<x;
}
else
return;
}
int main()
{
func(5);
return 0;
}

Answer 1

01234. Is the output after fixing semicolon.
The function call can be shown as :
fun(5) - > fun(4) - >fun(3) - > fun(2)-> fun(1) - > fun(0) - > return
The function are called in this way. We have a stack of function call which means that we return from func(0) then 1, then 2 and so on.

Base case: x <= 0

So function call terminates/ returns at fun(0)
Then we backtrack to the position which called func (0) i.e func(1) displaying 0
func (2) displaying 1
fun(3) displaying 2.. all the way to func(5) displaying 4.

The numbers are printed as 01234 because the cout statement is after each function call.

Answer 2

This is a recursive function.

Output should be something like:

01234

Why not run the code to check?

semicolon missing after func(x--)

Answer 3

A compile error because you are missing the ; after func (--x). Once fixed, it outputs 01234. func is called with 5, 4, 3, 2, 1, and 0. 0 just returns. The rest decrements x, calls the function, and prints x so the call with 1 gets to be the first to output it's 0.

Answer 4

VapeHorization plzz elaborate

Answer 5

You don’t need return because the type of function is Void.

If argument named x is greater than x, decrement and then print it. If not, return ( mistake )

Answer 6

Oh sorry I believed it was if (x<5), my mistake ^^

Answer 7

Xan bro I know the answer but I want to know how this program executed