Python tutorial says that (https://docs.python.org/3/tutorial/classes.html#python-scopes-and-namespaces)
In fact, local variables are already determined statically.
How to understand this? Based on what I knew static means that the type of a variable is determined at compile time. But it is not true considering for example
x = 1
x = 'str'
where the variable x is dynamically bound to objects of type int or string at runtime.
Reference: Is Python strongly typed?
In addition to the other answer, consider the error produced by the following code.
x = 1
def function():
y = x + 1
x = 3
function()
This will produce an error like "UnboundLocalError: local variable 'x' referenced before assignment" because it is determined that x is a local variable in function so it should be found in the local scope, negating the global definition.
x = x + 1, the lhs x causes a new local variable x to be created in this local scope. And as a result, the rls x would refer to this local x, instead of searching all the way to the global scope.
x + 1 without assignment thenTheir existence, and whether a variable lookup is local or global, is determined at compile time.
There is a statement in that document too.
if not declared nonlocal, those variables are read-only (an attempt to write to such a variable will simply create a new local variable in the innermost scope, leaving the identically named outer variable unchanged).
In Jared Goguen's code, clause x = 3 will let Python see x as local during compile.
Python: just names and objectsPythonhas no variables like most people used to think of them