Unable to bind local var to nested global var

Question

Unable to bind the local x of f() to the global var x of the nested g(). Why?

def f():
 x=0
 def g():
  global x
  x+=1
  print(x)
 g()
 g() # added to make seemingly more practical

-

>>> f()
...
NameError: global name 'x' is not defined

`x` is not a global variable, it is local to the function `f`. — user2390182, Sep 08 '18 at 16:43

Austin · Accepted Answer · 2018-09-08T17:05:49.443

3

You want to make x a global variable in function f():

def f():
    global x
    x = 0
    def g():
        global x
        x += 1
        print(x)
    g()

f()
# 1

As mentioned in comment, the 'global' state of x is pointless here, so it's better you pass x as argument to g() like so:

def f():
    x = 0
    def g(x):
        x += 1
        print(x)
    g(x)

f()
# 1

This not only makes your code more concise but also removes the overhead of 'global's.

edited Sep 08 '18 at 17:05

answered Sep 08 '18 at 16:44

Austin

One should mention, however, that the global variable in this code is utterly pointless. This function will always print `1`, no matter the "global" state of `x` – user2390182 Sep 08 '18 at 16:46
i also think it's genraly good practice not to use global, unless it's necesary – Jonas Wolff Sep 08 '18 at 17:09
1

@JonasWolff Ofcourse. Here is more details to it: https://stackoverflow.com/questions/19158339/why-are-global-variables-evil – Austin Sep 08 '18 at 17:11

score 0 · Answer 2 · edited Sep 08 '18 at 16:53

0

To avoid using globals, simply pass a parameter through the function g:

def f():
    x=0
    def g(y):
        y+=1
        print(y)
    g(x)

f()

This should work.

edited Sep 08 '18 at 16:53

Chandella07

answered Sep 08 '18 at 16:47

Oqhax

2 Answers2