passing same argument from outer function to inner function in python

Question

I have a inner function function

def inner_function(scenario = 'name',myopic_variable = True):

    if scenario == 'name':
        print('Name')

    if myopic_variable == True:
        print('Myopic is True')

and the main function like this.

def main_function(scenario = 'name',myopic_variable = True):
    inner_function()

Now when I am calling main function like this it works fine

main_function(scenario = 'name',myopic_variable = True)

but when i change myopic_variable = False it still gives me the same output why?

Answer 1

def inner_function(scenario = 'name',myopic_variable = True): You are passing a default value of true here.

def main_function(scenario = 'name',myopic_variable = True):
    inner_function()

Here, no parameters are passed to inner_functions() , it prints True , the default value. To make it correct, call inner_function(myopic_variable = myopic_variable)

Answer 2

Well that's because you are using something called default arguments in python. To understand what you are facing, let's take an example:

def inner_fun(a=True):
     print(a)

inner_fun() 

will print True because we are not passing any argument while calling this function. It will thus takr the default argument which is True here.

If you run inner_fun(False) , it'll print False because this time you are passing an argument in the function which has higher precedence than the default value.