Return not exiting function??

[rudihammad]

Hello,
I have this simple example where I simple want to exit a function when i == 5.

def foo(i):
    print i, "--"
    if i < 5:
        foo(i+1)
        print "++++++"
    elif i == 5:
        print "xxxxxx"
        return "what ever"
        
    print "why is this printing?"
    
x = foo(3)
print x

# Result:
3 --
4 --
5 --
xxxxxx
++++++
why is this printing?
++++++
why is this printing?
None <-----

2 things I don't get. 1-why is it not exiting the function? The fact that it is printing the last line shows that it is not exiting after return right? 2-(which I guess is derived from 1), why x is None when the return is "what ever"
I always use return to exit, but I an not sure what is happening here.

thx

[buran]

you call the function with argument i==3. It execute the if part (i.e. prints ++++++), because you call the function recursively when you call foo with i==5, it returns the what ever. You just throw it away and then it continues on line 10 and prints "why is this printing?" and then it exit the function (and return default None, because there is no explicit return.
It NEVER enters the elif block and never execute line 8 (where the only [explicit] return in the function is).

---

Try this site http://www.pythontutor.com/visualize.html
to visualise the execution and get better understanding
And by the way, don't use python2, use python3 instead. python2 is dead.

[rudihammad]

okey, I see, so it exiting the recursion and continues with the rest. Got it, that was silly
Thank you

[bowlofred]

You are calling the function more than once.

Sometimes when you run the function, the elif is taken, the "xxxxx" prints, and the function exits.
Then another time when the function is called the elif is skipped and the "why is this printing" appears.