Thread: Recursions

Hello, i took a test and had a problem with a function. I was wondering if someone could possibly explain the correct way of finding the answer. I have a final exam on this tomorrow and am still very confused! thanks for the help guys

Code:

for the function of p below, give the output of the call p(5)
void p( int x)
{
if (x<=0) return;
if (x%2==0) cout <<x<<endl;
p(x-1);
if (x%2 !=0) cout x<<endl;
}


for the function p below give the output of the call p(3)

void p (int x)
{
cout <<x<<endl;
if (x> 0) {p(x-1);p(x-1);}
}

for the first one i got only 5 but itshould be 4 2 1 3 5 and i have no idea how to figure that out

the second one should be 321001002100100

if any of you can explain how this concept works i would greatly appreciate it

Just follow the flow of the program and write down each step.

I'll start with the first one.

Call p with x = 5
If x < 0 return (5 < 0 is false so don't return)
If x &#37; 2 == 0 then output x (5 % 2 is 1 so don't output 5)
Call p with new x = current x - 1 (in other words call p with x = 4)
If x < 0 return (4 < 0 is false so don't return)
If x % 2 == 0 then output x (4 % 2 is 0 so output 4)
... and so on

i am starting to understand the first part of it going from 4 3 ect i dont understand how it goes back up? i figured once u got to a lower number like that it cant have an output of 5 at the end. the second one i am not understanding at all.. the first one i am seeing it more clearly

It goes back up as the call stack unwinds because your function prints old values of x before it calls p again and also before it returns.

>> i dont understand how it goes back up?
Post your step by step. What happens after you execute a return?

>> the second one i am not understanding at all..
Post your step by step like I did. It will be easier to help you understand where you are confused.

i will help a little since i can understand iterative ways better than recursive ones
#1

Code:

void p( int x) {
    while (x>0) {
        if (x&#37;2 == 0) cout << x << endl;
        --x;
        if (x%2 != 0) cout << x << endl;
    }
}