# defining 3 variables in global scope
my_var1, my_var2, my_var3 = 10, 20, 30

def some_fun():
  # declaring 'my_var1' as global inside a local scope,
  # so now 'my_var1' can be modified for global scope
  global my_var1

  # accessing a global scope variable, works fine
  print(my_var3)
  # accessing a global scope variable defined as global, works fine
  print(my_var1)
  # but the same thing with 'my_var2' doesn't work, why?
  # Note: comment below line to execute the program further
  print(my_var2)
  # Its because Python tries to figure out the scope of a variable by
  # first seeing if it has some assignment in this scope if it does Python
  # thinks it is a local variable, which we cannot access before
  # assignment right? In below lines 'my_var2' has assignment,
  # so it is considered as a local variable, even if same named global
  # variable exists it will be considered as a local variable
  # if the assignment is removed, 'my_var2' it will be considered global
  # and it can be accessed normally

  # modifying for global scope
  my_var1 = 30
  # modifying only for local scope
  my_var2 = 40

# now calling the function
some_fun()
# let's check our global variables
print(my_var1, my_var2)  # 30 \
                       # 10 \
                       # UnboundLocalError
# Notice: 'my_var1' is now changed but 'my_var2' is not

# similar to our previous example
def some_fun():
  # defining 'my_var1' and 'my_var2' in local scope
  my_var1 = 10
  my_var2 = 20

  def some_nested_fun():
      # declaring 'my_var2' as nonlocal, so now it can be modified for
      # 'some_fun()' scope too
      nonlocal my_var2
      # modifying for 'some_nested_fun()' scope
      my_var1 = 30
      # modifying for 'some_fun()' scope
      my_var2 = 40
      print(my_var1, my_var2)

  # calling the nested function
  some_nested_fun()
  # Notice: Here 'my_var1' is not modified but 'my_var2' is
  print(my_var1, my_var2)


# calling the outer function
some_fun() # 30 40 \
         # 10 40