Reputation: 1187
def scope_test():
def do_local():
spam = "local spam"
def do_nonlocal():
nonlocal spam
spam = "nonlocal spam"
def do_global():
global spam
spam = "global spam"
spam = "test spam"
do_local()
print("After local assignment:", spam)
do_nonlocal()
print("After nonlocal assignment:", spam)
do_global()
print("After global assignment:", spam)
scope_test()
print("In global scope:", spam)
Hi I am dealing with python class and faced with above example from the official documentation.
Output:
After local assignment: test spam
After nonlocal assignment: nonlocal spam
After global assignment: nonlocal spam
In global scope: global spam
The result of the first print after do_local
still prints "test spam" but can't figure out why while the second print gives "nonlocal spam".
What makes the difference?
My inference is, if I do_local() then it runs the do_local() and it changes the spam variable as "local spam" and it might have to work same with do_nonlocal() which results in "nonlocal spam". But it's not.
Why?
Upvotes: 2
Views: 256
Reputation: 4602
Here is a global variable example(from here):
globvar = 0
def set_globvar_to_one():
global globvar # Needed to modify global copy of globvar
globvar = 1
def print_globvar():
print(globvar) # No need for global declaration to read value of globvar
set_globvar_to_one()
print_globvar() # Prints 1
As you can see in this example, global means the variable which is available in any place in the script. If you define or change a global variable in a function, you can use it in another function with global keyword.
Even you can delete this first definiton line in the example:
globvar = 0
It works anyway as global variable. So, when you use global keyword it is available in any scope in the script.
nonlocal definition from here
Python 3 introduced the nonlocal keyword that allows you to assign to variables in an outer, but non-global, scope.
It means, when you use nonlocal keyword it is available outer scope. In this situation, "nonlocal spam" in do_nonlocal() function is available in scope_test() function too, but not one more outer scope.
If you don't use global and nonlocal keyword, then define "spam" in a function, it is just a local variable which is deleted when the function is over.
EDIT
This is not the same code, but prints the same result. For a better understanding, the situation is something like this:
spam_1 = ""
def scope_test():
def do_local():
spam_3 = "local spam"
def do_nonlocal():
nonlocal spam_2
spam_2 = "nonlocal spam"
def do_global():
global spam_1
spam_1 = "global spam"
spam_2 = "test spam"
do_local()
print("After local assignment:", spam_2)
do_nonlocal()
print("After nonlocal assignment:", spam_2)
do_global()
print("After global assignment:", spam_2)
scope_test()
print("In global scope:", spam_1)
Upvotes: 2