Python: variable bindings in recursive functions

Question

I came across some weirdness in Python using a function similar to the following:

    def foo(x):
        if int(x)!=4:
            x = raw_input("Wrong guess, please enter new value: " )
            foo(x)
        else:
            print "Good Job! %s was right"%x
        return x

    value = foo(x=raw_input("Guess a Number between 1 and 10: "))
    print value

If I enter, for instance: "1" then "2" then "3" then "4", I get the following printed out:

Good Job! 4 was right
2

This is confusing, since it seems the function is successfully identifying the right answer, but after doing so it is returning a value that was the 2nd response given, instead of the most recent response.

Can anyone explain what's going on with the binding of "x" in this recursive function?

Answer 1

Let's see!

value = foo(raw_input())
# foo is the same as in the question, I won't repeat it here
print value

Inside your foo, you get this:

# foo(1) calls foo(2) and sets x to 2
# foo(2) calls foo(3) and sets x to 3
# foo(3) calls foo(4) and sets x to 4
# foo(4) does:
print "Good Job! 4 was right"
return 4 # to foo(3)
# foo(3) does:
return 4 # to foo(2)
# foo(2) does:
return 3 # to foo(1)
# foo(1) does:
return 2 # to main

Since the return value to main (from the outermost recursion) is 2, that's what value remains.

To fix this, you can either make it iterative:

def iter_foo(x):
    while int(x) != 4:
        x = raw_input("Wrong guess. Try again! ")
    print "Good Job! %s was right" % x
    return x

Or make EACH recursion return the result of the new function

def recurse_foo(x):
    if int(x) != 4:
        return foo(raw_input("Wrong guess. Try again! "))
    else:
        print "Good Job! %s was right!" % x
        return x