Functions in Python - return, scope, args and kwargs

• Return in Function
• Scoping in Function
  • Access global variable in function context
  • Declare a local variable within in a function
  • Declare local variable with same name as in global
  • Using Global Keyword
• *args and **kwargs in function

Return in Function

Function execution stops once it reaches a return statement.

def add_num(x1,x2):
    return x1+x2
add_num(1,2)

Output:

3

The function will return 20, the addition expression will not execute as before that a return statement is reached.

def add_num(x1,x2):
    return 20
    return x1+x2
add_num(3,4)

Output:

20

Scoping in Function

• Access global variable in function context

Global variables can be accessed but can`t be altered. To altered the same, they has to be accessed by using global in front of them. We’ll see that later in the same section.

i = 20
def add_num(x1):
    return i+x1
add_num(2)

Output:

22

print(i)

Output:

20

• Declare a local variable within in a function

The scope of a local variable is limited to function itself. Outside the function it can`t be accessed and doing same will throw an error.

def add_num(x1):
    p = 40
    return p+x1

add_num(40)

Output:

80

Printing the local variable outside the function scope will throw an error.

try:
    print(p)
except Exception as e:
    print("Found Error - {0}".format(e))

Output:

Found Error - name 'p' is not defined

• Declare local variable with same name as in global

Local varibale will be given preference if a variable with sam name as the one outside the function is decalred but its scopre is limited in the function

i = 20
def add_num(x1):
    i =10
    return i+x1
add_num(4)

Output:

14

This will print the value of global variable.

print(i)

Output:

20

• Using Global Keyword

Global variable can`t be altered, to do the same access them via global keyword

i = 20
def add_num(x1):
    i = i -20
    return i+x1

try:
    add_num(5)
except Exception as e:
    print("Found Error - {0}".format(e))

Output:

Found Error - local variable 'i' referenced before assignment

i = 20
def add_num(x1):
    global i
    i = i -20
    return i+x1
add_num(5)

Output:

5

Accessing with global, will altered there value as well.

print(i)

Output:

0

args and kwargs in function

• args

User define arguments and keyword arguments (name, value pair) can be defined in function using *args and **kwargs. The *args have to be defined after positional argument and kwargs has to be defined after *args. There is a strict order is defined if any function incldue these all.

def print_name(*args):
    for argument in args:
        print(argument)

print_name("hello","world!")

Output:

hello
world!

• kwargs

def print_name(**kwargs):
    print(kwargs)
    print(kwargs.items())
    for item in kwargs:
        print(item, kwargs[item])
    for name, value in kwargs.items():
        print( '{0} = {1}'.format(name, value))

print_name(a = "hello",b="world!")

Output:

{'a': 'hello', 'b': 'world!'}
dict_items([('a', 'hello'), ('b', 'world!')])
a hello
b world!
a = hello
b = world!

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