In Python, variables are the containers for storing data values. They are reference, or pointers, to an object in memory which means that whenever a variable is assigned to an instance, it gets mapped to that instance. 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.
Example:
# Python program to demonstrate # variable assignment # An integer assignment age = 45 # A floating point salary = 1456.8 # A string name = "John" print(age) print(salary) print(name) |
Output:
45 1456.8 John
Note: To know more about variables click here.
Scope of variable
The location where we can find a variable and also access it if required is called the scope of a variable.
Global and local 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:
# This function uses global variable s def f(): print(s) # Global scope s = "I love Geeksforgeeks"f() |
Output:
I love Geeksforgeeks
Now suppose a variable with the same name is defined inside the scope of function as well then it will print the value given inside the function only and not the global value.
# This function has a variable with # name same as s. def f(): s = "Me too." print(s) # Global scope 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 is no local s, 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:
def f(): print(s) # This program will NOT show error # if we comment below line. s = "Me too." print(s) # Global scope s = "I love Geeksforgeeks" f() print(s) |
Output:
Traceback (most recent call last): File "/home/370cac45bae7f1e6096520b7a0edb604.py", line 13, inf() 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:
# This function modifies global variable 's' def f(): global(s) print(s) s = "Look for Geeksforgeeks Python Section" print(s) # Global Scope 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 better understanding of the topic.
# Python program to demonstrate # scope of variable a = 1 # Uses global because there is no local 'a' def f(): print('Inside f() : ', a) # Variable 'a' is redefined as a local def g(): a = 2 print('Inside g() : ', a) # Uses global keyword to modify global 'a' def h(): global a a = 3 print('Inside h() : ', a) # Global scope 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
Nonlocal keyword
In Python, nonlocal keyword is used in the case of nested functions. This keyword works similar to the global, but rather than global, this keyword declares a variable to point to the variable of outside enclosing function, in case of nested functions.
Example:
# Python program to demonstrate # nonlocal keyword print ("Value of a using nonlocal is : ", end ="") def outer(): a = 5 def inner(): nonlocal a a = 10 inner() print (a) outer() # demonstrating without non local # inner loop not changing the value of outer a # prints 5 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
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