How can I share a function variable within a nested function?

Question

UPDATED:

How can I use a function variable within nested function? I've simplified my problem in the following example:

def build():
    n = 1 # code to parse rpm package minor version from file
    f = min_ver(n) # update minor version
    return

def min_ver(n):
    n = 2 # this is defined by another process, not set intentionally
    s =  1 + n # still need the original value from build()
    return s

The actual use case is that I'm grabbing a parsed rpm package minor version value in ex1() from disk called 'n'. When ex2() is executed from ex1(), it deletes the old package, builds a new rpm package with a new minor version. So when it calls for ex1()'s value within the nested function, it's not changed to the new version.

How can I maintain the original 'n' value within the nested function, before passing onto a new value of 'n' post nested function?

Nothing is going to change the local scope n within ex1 in the background. If you declare n as a global within ex1 then something else could change it externally. — B. Morris
– B. Morris, Commented Nov 29, 2018 at 2:05

DMe · Accepted Answer · 2018-11-29 02:35:50Z

2

A simple way to do this would be to pass the variable as an argument to ex2.

def build():
    n = int(1)
    f = ex2(n) # pass the value to the next function
    n = int(5) 
    return 

def min_ver(n_old):
    n = 2
    s =  1 + n_old # use the n that was passed in
    return s

edited Nov 29, 2018 at 2:35

answered Nov 29, 2018 at 1:51

DMe

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4 Comments

DMe Over a year ago

Can you not call ex2 before the background task changes n instead? Otherwise you could save the old n to a new variable before it is changed.

DMe Over a year ago

As in my updated answer - if you name the argument to min_ver something else, then it will retain that value even if you change n.

hobbes Over a year ago

Somethings I over complicate things in my head, thank you.

DMe Over a year ago

Glad I could help, happens to all of us!

B. Morris · Accepted Answer · 2018-11-29 02:03:55Z

0

If you make ex2() actually nested then you can access the outer variables.

def ex1():
    n = int(1)
    def ex2():
        s =  1 + n
        return(s)
    f = ex2()
    n = int(5) # done with nested value, rewrite new value
    return()

Also, you probably want to return f or n instead of an empty tuple, I would imagine.

And you don't need to say int(1) you can just say 1. Everything, including integers and strings, is implicitly an object in python.

answered Nov 29, 2018 at 2:03

B. Morris

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