Python scope is prone to causing variable errors, and the key is to understand the use of local, global and nonlocal. 1. The variable assigned with = in the function is local by default, even if the same name as the global variable does not affect each other; 2. Only external variables are read without declaration, but global variables are required to be declared when modifying, otherwise local variables will be created; 3. Nonlocal is required to modify outer non-global variables in nested functions, indicating that the most recent outer variable is used, and cannot be cross-level or used for global variables. Mastering these rules can effectively avoid variable reference errors.
Python's scope looks simple, but many people will encounter problems such as variables not being found or assignment errors when actually writing code. The key is to figure out when and which variable is used, especially the usage scenarios of local
, global
and nonlocal
keywords.

The variable defined inside the function is local
When you use =
to assign a value to a variable in a function, the variable is a local variable by default. Even if you define a variable with the same name outside the function, the global variable will not be automatically used inside the function.
For example:

x = 10 def func(): x = 5 print(x) func() # Output 5 print(x) # Output 10
At this time, x
in the function and x
outside are two different variables. If you just read the external variable without modifying it, it is accessible:
x = 10 def read_x(): print(x) read_x() # Output 10
But as long as you try to assign a value to x
in a function, such as x = ...
, Python will treat it as a local variable unless you specifically specify it.

Want to change global variables in a function? Use global
If you really want to change the value of a global variable in a function, you must declare it global
:
x = 10 def change_global(): global x x = 20 change_global() print(x) # Output 20
If global
is not added, it will become the situation mentioned above: a local variable x
is created in the function, which will not affect the outside.
Common errors are written like this:
x = 10 def bad_func(): x = 1 # An error will be reported here UnboundLocalError
Because x = 1
is equivalent to x = x 1
, and global x
is not declared in the function, Python thinks that you are operating on local variables, but you have not assigned a value at this time, and you have an error.
Want to modify outer non-global variables in nested functions? Use nonlocal
If there are functions (nested functions) in the function, if you want to modify the variables of the outer function, you must use nonlocal
.
See an example:
def outer(): x = 10 def inner(): nonlocal x x = 20 inner() print(x) # Output 20 outer()
Here, nonlocal x
is used in inner()
, telling Python that I want to use the x
closest to me that is not global, that is, the one in outer()
. If nonlocal
is not added, then x = 20
will create a new local variable in inner()
, which has no effect on the outer layer.
A few points to note:
-
nonlocal
can only be used in nested functions and cannot be used to modify global variables - The referenced variable must already exist in the outer scope and cannot be found "cross-level".
- If there is no corresponding variable in the outer layer, an error will be reported
Basically that's it. Understanding the difference between local, global, and nonlocal can avoid many inexplicable bugs. Especially when writing closures or multi-layer nested functions, remember to check whether the variable is the one you want.
The above is the detailed content of Understanding Python scope: local, global, nonlocal. For more information, please follow other related articles on the PHP Chinese website!

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