Optional arguments in nested functions in Python

Question

Is it possible for one function to take as an argument the name of one of the optional arguments of a second function then call the second function with the optional argument set to the value of some variable in the first function?

Code (that obviously doesn't work)

def foo(a=1,b=2,c=3):
    d = a + b + c
    return d

def bar(variable):
    z = get_user_input()
    e = foo(variable = z)
    return e

print(bar(a))

The desired result is for bar to call foo(a=z) and print whatever z+2+3 is. Of course Python doesn't know what (a) is here. My guess is that I can somehow reference the list of arguments of foo() , but I am stumped as to how you might do that.

Do you mean to call bar with the name of the variable as a string: print(bar('a'))? Your code currently would raise an error — Iain Shelvington
– Iain Shelvington, Commented Oct 13, 2021 at 2:57

Nguyễn Minh Hiếu · Accepted Answer · 2021-10-13 03:00:16Z

1

Maybe try the code snippet below

def foo(a=1,b=2,c=3):
    d = a + b + c
    return d

def bar(variable: str):
    z = int(input())
    e = foo(**{variable: z})
    return e

# variable name should be defined as string
print(bar("a"))

The ** parses all arbitrary arguments on dict, in this case is a. Be careful tho, as if passing wrong variable name (differ from a, b or c) will result raising error.

answered Oct 13, 2021 at 3:00

Nguyễn Minh Hiếu

6771 gold badge5 silver badges12 bronze badges

Sign up to request clarification or add additional context in comments.

1 Comment

Lemming Over a year ago

Exactly what I was looking for. Thank you!

Collectives™ on Stack Overflow

Optional arguments in nested functions in Python

1 Answer 1

1 Comment

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

1 Answer 1

1 Comment

Your Answer

Sign up or log in

Post as a guest

Related