Using Global Variables inside a Nested Function in Python

Question

I read this code (given below) and my understanding was that, if a variable is declared global inside a function and if it is modified then it's value will change permanently.

x = 15
def change(): 
    global x 
    x = x + 5
    print("Value of x inside a function :", x) 
change() 
print("Value of x outside a function :", x)

Output:

Value of x inside a function : 20
Value of x outside a function : 20

But the code below shows a different output. How is it that the value of x does not change inside the print("After making change: ", x) and still remains 15

def add(): 
    x = 15
    
    def change(): 
        global x 
        x = 20
    print("Before making changes: ", x) 
    print("Making change") 
    change() 
    print("After making change: ", x) 

add() 
print("value of x",x)

Output:

Before making changes:  15
Making change
After making change:  15
value of x 20

add does not read a global x, it reads the local x created by x = 15. add would have to declare x as global as well. — MisterMiyagi
– MisterMiyagi, Commented Jul 10, 2020 at 16:05
But wouldn't the value of x change permanently when the change() function is called ? — Mihir
– Mihir, Commented Jul 10, 2020 at 16:15

chepner · Accepted Answer · 2020-07-10 16:20:40Z

12

In add, x is not a global variable; it's local to add. You either need to make it global as well, so that add and change are referring to the same variable

def add(): 
    global x
    x = 15
    
    def change(): 
        global x 
        x = 20
    print("Before making changes: ", x) 
    print("Making change") 
    change() 
    print("After making change: ", x) 

add() 
print("value of x",x)

or you need to declare x in change as nonlocal, rather than global.

def add(): 
    x = 15
    
    def change(): 
        nonlocal x 
        x = 20
    print("Before making changes: ", x) 
    print("Making change") 
    change() 
    print("After making change: ", x) 

add() 
print("value of x",x)

answered Jul 10, 2020 at 16:20

chepner

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3 Comments

velenos14 Over a year ago

in the second box, the last print("value of x", x) produces a NameError: name 'x' is not defined. pretty understandable as the only x's which come into play are those from add(), i.e. local to add(), and the x from change(), which is preceded by nonlocal, thus referring to the x from add. there's no ''global'' x, thus the print from outside all functions doesn't actually have what to print after the add() function call has finished and the stack has been emptied from any x's. correct?

chepner Over a year ago

Yeah, looks like I incorrectly removed the original global declaration from add. Putting it back, I'm not sure there's any difference between global and nonlocal in change, or any reason to use nonlocal instead, so I'm not entirely sure what my point there was. (Probably just blanked on what I was doing in my haste to talk about nonlocal.)

o17t H1H' S'k Over a year ago

the second option wont work, and you need to use global in both functions. with global + nonlocal you get a SyntaxError: no binding for nonlocal 'x' found

Vishal Singh · Accepted Answer · 2020-07-10 16:31:42Z

Inside the function add the variable x is present in the enclosing scope and not the global scope that's why your variable x won't change even if you use the global keyword.
To change this x you need nonlocal keyword which applies to the variables present in the enclosing scope.

The nonlocal statement causes the listed identifiers to refer to previously bound variables in the nearest enclosing scope excluding globals.


# global scope
x = 0  


def add():
    # enclosing scope
    x = 15

    def change():
        nonlocal x
        x = 20

    print("Before making changes: ", x)
    print("Making change")
    change()
    print("After making change: ", x)


add()
print("value of x", x)  # this x is global variable

Output:

Before making changes:  15
Making change
After making change:  20
value of x 0

Javad Rezaei · Accepted Answer · 2020-07-10 16:15:56Z

1

when you define global x in change() function, nested in add() function; it defines a main global x variable which is different from local x = 15 at the beginning of the add() function. when you print x before and after calling change(), you actually use local x = 15 at the beginning of add() function, but final print after calling add() will use defined global x which defined in main scope and has the value of 20

more explanation

answered Jul 10, 2020 at 16:15

Javad Rezaei

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