Python Recursive Function Call With If Statement

I have a question regarding function-calls using if-statements and recursion. I am a bit confused because python seems to jump into the if statements block even if my function retu

Solution 1:

Short answer:

That's because you're not actually calling the function. You can call the function by using the parenthesis.

if function2():
    ...

Long answer:

Functions in Python are a first class citizen (functional paradigm), and so it is completely valid to refer to a function just by its name. The following is valid syntax:

def hello():
    print("Hello")

hello_sayer =hello
hello_sayer()# print "Hello"

The next concept in play is truth-ness of non-Boolean variables. In Python, the following are considered False-y

• None
• False
• zero of any numeric type, for example, 0, 0L, 0.0, 0j.
• any empty sequence, for example, '', (), [].
• any empty mapping, for example, {}. instances of user-defined classes, if the class defines
• a nonzero() or len() method, when that method returns the integer zero or bool value False.

Everything else is True-ish. Since a function name falls under none of the above categories, it is considered True-ish when tested in a conditional context.

Reference: https://docs.python.org/3/library/stdtypes.html#truth-value-testing

Edit: The earlier question was incomplete, and did not have a function call. For the new question, AChampion's answer is the correct one.

Solution 2:

You may misunderstand how recursion works, yes it continues at line 5 or 6 because the recursion has ended at a lower level in the call stack, so it continues at a higher-level in the call stack. Here's a sample call stack, note the next operation after False is the next findExit() at the higher call stack:

1 findExit(...):
2True:
3        field assignment
4.1      findExit(x+1) 
  2True3              field assignment
  4.1            findExit(x+1):
    2False# Does not jump to line 5 in current call stack.5.1            findExit(x-1):
    .                ...