Re: amazing scope?
- 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..
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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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Re: amazing scope?
- Original Message -
2012/11/30 andrea crotti andrea.crott...@gmail.com:
Already changing it to:
def record_things():
out.write(Hello world)
def main():
with open('output', 'w') as out:
record_things()
if __name__ == '__main__':
main()
makes it stops working as expected, so it's really just a corner case
of using the if __name__ == '__main__'
which I had never encountered before..
You do realize that
foo = 5 # define at the module level
if __name__ == '__main__':
bar = 6 # also at the module level
if True:
ham = 8 # still at the module level
def func()
# scope has changed, only at the func level
jam = 9
Nothing magic about if __name__ == '__main__'
Cheers,
JM
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Re: amazing scope?
Am 30.11.2012 12:11, schrieb andrea crotti:
I wrote a script, refactored it and then introducing a bug as below:
def record_things():
out.write(Hello world)
This is a function. Since out is not a local variable, it is looked up
in the surrounding namespace at the time the function is called.
if __name__ == '__main__':
with open('output', 'w') as out:
record_things()
This is not in a function, so binding variables affects the containing
namespace. This includes binding the context manager to out. Note that
you could have moved the whole code into a function (e.g. one called
main()) and then you would have gotten the expected failure.
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..
Other than in C/C++/Java and others, indention doesn't introduce a new
scope, but I understand your intuition. Even simpler, this is how my
early Python code looked like, in the absence of the C ternary operator
and the distinction between a variable declaration and an assignment:
foo = None
if some_condition:
foo = 'bar'
else:
foo = 'baz'
More idiomatic would have been this:
if some_condition:
foo = 'bar'
else:
foo = 'baz'
Summary: I'd say that everything is fine. ;)
Uli
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Re: amazing scope?
On Sat, Dec 1, 2012 at 3:05 AM, andrea crotti andrea.crott...@gmail.com wrote:
Well I knew that this works fine, even if I feel a bit guilty to do
this, and better is:
foo = 'bar' if some_condition else 'baz'
Anyway for me the suprise is that something that is defined *later* at
the module scope is found in a function which is defined *earlier*.
It's assigned to earlier. It's no different from C code like this:
int foo;
void blah()
{
/* do stuff with foo */
}
int main()
{
foo = 1;
}
In Python, the int foo; is implicit, but the scope is the same.
ChrisA
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Re: amazing scope?
- Original Message - Well I knew that this works fine, even if I feel a bit guilty to do this, and better is: foo = 'bar' if some_condition else 'baz' Anyway for me the suprise is that something that is defined *later* at the module scope is found in a function which is defined *earlier*. -- http://mail.python.org/mailman/listinfo/python-list What really matters for function is when they're executed, not defined (except for their parameter default value). foo = 'I am foo' bar = 'I am bar' def funcA(param1=foo): print param1 print bar foo = 'I am another foo' bar = 'I am another bar' def funcB(param1=foo): print param1 print bar funcA() funcB() I am foo I am another bar I am another foo I am another bar cheers, JM -- IMPORTANT NOTICE: The contents of this email and any attachments are confidential and may also be privileged. If you are not the intended recipient, please notify the sender immediately and do not disclose the contents to any other person, use it for any purpose, or store or copy the information in any medium. Thank you. -- http://mail.python.org/mailman/listinfo/python-list
RE: amazing scope?
andrea crotti
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..
Makes sense to me. `out` is declared in an if statement. If statements
have no scope and it is not in a function so it gets added to the
module's namespace.
~Ramit
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Re: amazing scope?
On 11/30/2012 11:05 AM, andrea crotti wrote: Well I knew that this works fine, even if I feel a bit guilty to do this, and better is: foo = 'bar' if some_condition else 'baz' Anyway for me the suprise is that something that is defined *later* at the module scope is found in a function which is defined *earlier*. It would have been nice if you had indicated in your original post just *what* you considered shocking about the code sample. The code was valid, and we all had to guess what about it was puzzling you. Like most others, I first guessed you were confused about the definition being in an if statement. Then I thought you were confused by the with statement, since it's not as obvious that as binds the name to the object in the same way as assignment. But I never would have guessed that you were confused about the order of execution in a module. The source code in a module is executed in order, including function definitions defined at top level. However, such a function definition's execution creates a function object, and assigns a global name (usually) to that object. It does not execute the object. When that function is executed later, it then looks for global variables for stuff that's not defined within its own scope. So by the time the function references the 'out' name, it's been successfully added to the global namespace. -- DaveA -- http://mail.python.org/mailman/listinfo/python-list
