Python Scope of Variables
In Python, variables are the containers for storing data values. Unlike other languages like C/C++/JAVA, Python is not “statically typed”. We do not need to declare variables before using them or declare their type. A variable is created the moment we first assign a value to it.
Python Scope variable
The location where we can find a variable and also access it if required is called the scope of a variable.
Python Local variable
Local variables are those that are initialized within a function and are unique to that function. It cannot be accessed outside of the function. Let’s look at how to make a local variable.
Python3
def f():
s = "I love Geeksforgeeks"
print (s)
f()
|
Output
I love Geeksforgeeks
If we will try to use this local variable outside the function then let’s see what will happen.
Python3
def f():
s = "I love Geeksforgeeks"
print ( "Inside Function:" , s)
f()
print (s)
|
Output:
NameError: name 's' is not defined
Python Global variables
Global variables are the ones that are defined and declared outside any function and are not specified to any function. They can be used by any part of the program.
Example:
Python3
def f():
print (s)
s = "I love Geeksforgeeks"
f()
|
Output:
I love Geeksforgeeks
Global and Local Variables with the Same Name
Now suppose a variable with the same name is defined inside the scope of the function as well then it will print the value given inside the function only and not the global value.
Python3
def f():
s = "Me too."
print (s)
s = "I love Geeksforgeeks"
f()
print (s)
|
Output:
Me too.
I love Geeksforgeeks
The variable s is defined as the string “I love Geeksforgeeks”, before we call the function f(). The only statement in f() is the print(s) statement. As there are no locals, the value from the global s will be used.
The question is, what will happen if we change the value of s inside of the function f()? Will it affect the global s as well? We test it in the following piece of code:
Python3
def f():
print (s)
s = "Me too."
print (s)
s = "I love Geeksforgeeks"
f()
print (s)
|
Output:
Traceback (most recent call last):
File "/home/370cac45bae7f1e6096520b7a0edb604.py", line 13, in
f()
File "/home/370cac45bae7f1e6096520b7a0edb604.py", line 3, in f
print(s)
UnboundLocalError: local variable 's' referenced before assignment
To make the above program work, we need to use global keyword. We only need to use global keyword in a function if we want to do assignments / change them. global is not needed for printing and accessing. Why? Python “assumes” that we want a local variable due to the assignment to s inside of f(), so the first print statement throws this error message. Any variable which is changed or created inside of a function is local, if it hasn’t been declared as a global variable.
To tell Python, that we want to use the global variable, we have to use the keyword global, as can be seen in the following
Example:
Python3
def f():
global s
print (s)
s = "Look for Geeksforgeeks Python Section"
print (s)
s = "Python is great !"
f()
print (s)
|
Output:
Python is great!
Look for Geeksforgeeks Python Section
Look for Geeksforgeeks Python Section
Consider the below example for a better understanding of the topic.
Python3
a = 1
def f():
print ( 'Inside f() : ' , a)
def g():
a = 2
print ( 'Inside g() : ' , a)
def h():
global a
a = 3
print ( 'Inside h() : ' , a)
print ( 'global : ' , a)
f()
print ( 'global : ' , a)
g()
print ( 'global : ' , a)
h()
print ( 'global : ' , a)
|
Output:
global : 1
Inside f() : 1
global : 1
Inside g() : 2
global : 1
Inside h() : 3
global : 3
Python Nonlocal keyword
In Python, the nonlocal keyword is used in the case of nested functions. This keyword works similarly to the global, but rather than global, this keyword declares a variable to point to the variable of an outside enclosing function, in case of nested functions.
Example:
Python3
print ( "Value of a using nonlocal is : " , end = "")
def outer():
a = 5
def inner():
nonlocal a
a = 10
inner()
print (a)
outer()
print ( "Value of a without using nonlocal is : " , end = "")
def outer():
a = 5
def inner():
a = 10
inner()
print (a)
outer()
|
Output:
Value of a using nonlocal is : 10
Value of a without using nonlocal is : 5
Last Updated :
23 Mar, 2023
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