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Yes, I stumbled across this quite recently: https://x.com/marekgibney/status/1817495719520940236 https://x.com/marekgibney/status/1817495719520940236 I was su

by mg 2y ago

Yes, I stumbled across this quite recently: https://x.com/marekgibney/status/1817495719520940236 https://x.com/marekgibney/status/1817495719520940236 I was surprised that you cannot access dynamically created local variables even though they exist. Looks like the local dictionary still exists. Just that whether accessing a local variable looks it up in the local or global dictionary is determined at compile time.

eesmith 2y ago
See PEP 667, "Consistent views of namespaces" https://peps.python.org/pep-0667/ https://peps.python.org/pep-0667/ . In Python 3.12, quoting https://docs.python.org/dev/whatsnew/3.13.html https://docs.python.org/dev/whatsnew/3.13.html > PEP 667: The locals() builtin now has defined semantics when mutating the returned mapping. Python debuggers and similar tools may now more reliably update local variables in optimized scopes even during concurrent code execution. with more details at https://docs.python.org/dev/whatsnew/3.13.html#whatsnew313-locals-semantics https://docs.python.org/dev/whatsnew/3.13.html#whatsnew313-l... , which adds: > To ensure debuggers and similar tools can reliably update local variables in scopes affected by this change, FrameType.f_locals now returns a write-through proxy to the frame’s local and locally referenced nonlocal variables in these scopes
- mg 2y ago
  Interesting. As I understand it, it will not affect my example: def f(): exec('y = 7') # will still create a local variable print(locals()['y']) # will still print this local variable print(y) # will still try to access a global variable 'y'
  - eesmith 2y ago
    That does not work for me in Python 3.10 or 3.12: Python 3.12.2 ... Type "help", "copyright", "credits" or "license" for more information. >>> def f(): ... exec('y = 7') # will still create a local variable ... print(locals()['y']) # will still print this local variable ... print(y) # will still try to access a global variable 'y' ... >>> f() 7 Traceback (most recent call last): File "<stdin>", line 1, in <module> File "<stdin>", line 4, in f NameError: name 'y' is not defined It does print 7 in Python 2.7.
    - mg 2y ago
      That's what I meant. It tries to access a global variable 'y' and fails because there is none. See also: https://x.com/marekgibney/status/1817495719520940236 https://x.com/marekgibney/status/1817495719520940236 and https://x.com/marekgibney/status/1817495721974583569 https://x.com/marekgibney/status/1817495721974583569
      - deleted 2y ago
        [deleted]
      - eesmith 2y ago
        I see what you mean now. Messing with locals has been a long-time danger zone. I don't know how it is supposed to interact with exec - another danger zone - and I don't know enough about the changes in 3.13 to know if it's resolved.

by mg 2y ago

eesmith 2y ago

See PEP 667, "Consistent views of namespaces" https://peps.python.org/pep-0667/ https://peps.python.org/pep-0667/ . In Python 3.12, quoting https://docs.python.org/dev/whatsnew/3.13.html https://docs.python.org/dev/whatsnew/3.13.html > PEP 667: The locals() builtin now has defined semantics when mutating the returned mapping. Python debuggers and similar tools may now more reliably update local variables in optimized scopes even during concurrent code execution. with more details at https://docs.python.org/dev/whatsnew/3.13.html#whatsnew313-locals-semantics https://docs.python.org/dev/whatsnew/3.13.html#whatsnew313-l... , which adds: > To ensure debuggers and similar tools can reliably update local variables in scopes affected by this change, FrameType.f_locals now returns a write-through proxy to the frame’s local and locally referenced nonlocal variables in these scopes

mg 2y ago
Interesting. As I understand it, it will not affect my example: def f(): exec('y = 7') # will still create a local variable print(locals()['y']) # will still print this local variable print(y) # will still try to access a global variable 'y'
- eesmith 2y ago
  That does not work for me in Python 3.10 or 3.12: Python 3.12.2 ... Type "help", "copyright", "credits" or "license" for more information. >>> def f(): ... exec('y = 7') # will still create a local variable ... print(locals()['y']) # will still print this local variable ... print(y) # will still try to access a global variable 'y' ... >>> f() 7 Traceback (most recent call last): File "<stdin>", line 1, in <module> File "<stdin>", line 4, in f NameError: name 'y' is not defined It does print 7 in Python 2.7.
  - mg 2y ago
    That's what I meant. It tries to access a global variable 'y' and fails because there is none. See also: https://x.com/marekgibney/status/1817495719520940236 https://x.com/marekgibney/status/1817495719520940236 and https://x.com/marekgibney/status/1817495721974583569 https://x.com/marekgibney/status/1817495721974583569
    - deleted 2y ago
      [deleted]
    - eesmith 2y ago
      I see what you mean now. Messing with locals has been a long-time danger zone. I don't know how it is supposed to interact with exec - another danger zone - and I don't know enough about the changes in 3.13 to know if it's resolved.