Question

I want to do something like this in Python. I took out what we are really doing and replaced it with this ridiculous example that will really never end. Please assume the calls from a<==>b are finite. Our code does have the logic to end the cycle.

I am worried that I will get an error for calling b in a before b is defined. However, I read that as long as I don’t make a call that executes a before def b then I should have no problem. What is the real answer here? And what is python doing behind the scenes to make it not exit on line 2 (b())

def a():
    b()

def b():
    a()

b()

Answer 1

The real answer is that b inside the definition of a will be looked up in the module scope and similarly for a inside the definition of b. Since a and b both exist in the module scope after both definitions have been processed, your mutual recursion will work.

(It will stop working if the names a and b are shadowed inside the function definitions, but I assume you’ll manage to avoid that.)

See this question for an overview of Python scoping rules.