redefinition in closure functions

[MirS0bhan]

Bug report

Bug description:

i think it could be a bug or maybe we just need a PEP but I issue it here
in closure functions we cant redefine a var that has defined before in the parent function
as the normal action I think it should resolve it because is happens normally but when we use it in a
redefinition it doesn't work.

def foo(x: int):
    x: int = x * x
    def john():
        x: int = x + x
        return x
    return john

error: Unresolved reference 'x'

CPython versions tested on:

3.12

Operating systems tested on:

Linux

[ZeroIntensity]

You can do this with the nonlocal keyword.

[MirS0bhan]

should not it be a norm behavior ?

[ZeroIntensity]

That would break untold amounts of code :(

You can refactor to use nonlocal to get the behavior you want:

def foo(x: int):
    x: int = x * x
    def john():
        nonlocal x
        x: int = x + x
        return x
    return john

[ZeroIntensity]

@encukou, this should get closed as not planned.

[terryjreedy]

Requiring nonlocal to modify nonlocal names is the same as requiring global to modify globals. Otherwise, names are looked up in locals first. The keyword says to skip locals.

[terryjreedy]

Questions like this should be posted to the help channel on discuss.python.org.

[MirS0bhan]

actually it wasn't just a question. I suggest we could automate it because that's so simple move. we had call it before of definition so it should work. anyway