Please tell me how does "nonlocal" work in Python?
Closed due to the fact that the issue is too general for the participants Sergey Gornostaev , Xander , Dmitry Kozlov , 0xdb , Air 19 Mar at 2:59 .
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1 answer
When declaring a variable through nonlocal, it will refer to a variable with the same name in the nearest closure, excluding global variables
For example, ad without nonlocal
x = 0 def outer(): x = 1 def inner(): x = 2 print("inner:", x) inner() print("outer:", x) outer() print("global:", x) # inner: 2 # outer: 1 # global: 0 And now with nonlocal, x in inner () is x in outer ()
x = 0 def outer(): x = 1 def inner(): nonlocal x x = 2 print("inner:", x) inner() print("outer:", x) outer() print("global:", x) # inner: 2 # outer: 2 # global: 0 And with global will be so
x = 0 def outer(): x = 1 def inner(): global x x = 2 print("inner:", x) inner() print("outer:", x) outer() print("global:", x) # inner: 2 # outer: 1 # global: 2 
Sergey Konovalov Sergey Konovalov
854 7 bronze marks
- Thank you very much everything is very clear and understandable - Alexey
- I apologize, it is a little incomprehensible how the global works - Alex
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