Use of nonlocal vs use of global keyword in Python

15 Jun, 2021
Prerequisites: Global and Local Variables in PythonBefore moving to nonlocal and global in Python. Let us consider some basic scenarios in nested functions.
Python3
def fun():    var1 = 10     def gun():        print(var1)        var2 = var1 + 1        print(var2)     gun()fun()
10
11
 
The variable var1 has scope across entire fun(). It will be accessible from nested function of fun()
Python3
def fun():    var1 = 10     def gun():        # gun() initializes a new variable var1.        var1 = 20        print(var1, id(var1))     print(var1, id(var1))    gun()fun()
10 10853920
20 10854240
 
In this case gun() initialized new variable var1 in the scope of gun. var1 with value 10 and var1 with value 20 are two different and unique variables. var1 holding value 20 will be by default accessed in gun().
Considering the previous example, we know that guns will initialize a new variable var1 in its own scope. But when it is going to do so, it cannot find the value of var1 yet, to perform the arithmetic operation as no value has been assigned to var1 previously in gun(). 
Python3
def fun():    var1 = 10     def gun():        # tell python explicitly that it        # has to access var1 initialized        # in fun on line 2        # using the keyword nonlocal        nonlocal var1        var1 = var1 + 10        print(var1)     gun()fun()
20
 
In this example, before initializing var1 in gun(). We have explicitly told Python that, do not initialize a new variable, instead access var1 present already on line 2. using the keyword nonlocal So when the interpreter performs addition, it access the value 10(already present) and error is avoided.
Let us move on to global keyword now. Consider the examples given below 
Python3
var1 = 10def fun():    # global variable var1 will# be read or accessed in fun()    print('var1 is', var1) fun()
var1 is 10
 
Global variable var1 will be simply read or accessed inside function fun()
Python3
var1 = 10def fun():    # new local variable var1    # will be initialized in fun()    var1 = 20    print('var1 is', var1)    print('var1 is at', id(var1)) fun()print('var1 is', var1)print('var1 is at', id(var1))
var1 is 20
var1 is at 10854240
var1 is 10
var1 is at 10853920
 
A new variable var1 will be initialized in fun(). Global variable var1 will be different from local variable var1 of fun(). By default in fun(), the local variables will be accessed.
Considering previous example, we know that fun() will initialize a new variable var1 in its own scope. But when it is going to do so, it cannot find the value of var1 yet, to perform arithmetic operation as no value has been assigned to var1 previously in fun()
Python3
var1 = 10def fun():    # tell python explicitly do not    # initialise a new variable    # instead access var1 which    # has global scope    global var1    var1 = var1 + 20    print('var1 is', var1) fun()
var1 is 30
 
In this example, before initializing var1 in fun(). We have explicitly told python that, do not initialize a new variable, instead access var1 present already on line 1. using the keyword global So when interpreter performs addition, it access the value 10(already present), and error is avoided. 
anikaseth98
python-basics
Python
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