----- Original Message -----
> I wrote a script, refactored it and then introducing a bug as below:
>
> def record_things():
> out.write("Hello world")
>
> if __name__ == '__main__':
> with open('output', 'w') as out:
> record_things()
>
>
> but the shocking thing is that it didn't actually stopped working, it
> still works perfectly!
>
> What my explanation might be is that the "out" is declared at module
> level somehow,
> but that's not really intuitive and looks wrong, and works both on
> Python 2.7 and 3.2..
> --
> http://mail.python.org/mailman/listinfo/python-list
>
>From what I understand from the with documentation
>http://docs.python.org/2/reference/compound_stmts.html#with
The with block has nothing to do with scopes. It does not define the target
within that block, well, actually it does but it defines it for the current
scope.
The with block only identifies where to call __enter__ and __exit__.
So I would say that "with open('output', 'w') as out" assignes "out" at the
module level.
You may have been mislead by some other languages where a with block purpose is
to limit the scope of a variable.
JM
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