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@dcrosta
Created April 30, 2012 20:17
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A Python Mystery
Guess the output:
>>> def exec_code_object_and_return_x(codeobj, x):
... exec codeobj
... return x
...
>>> co1 = compile("""x = x + 1""", '<string>', 'exec')
>>> co2 = compile("""del x""", '<string>', 'exec')
>>> exec_code_object_and_return_x(co1, 1)
# What do you get here?
>>> exec_code_object_and_return_x(co2, 1)
# And what about here?
@dcrosta
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dcrosta commented May 1, 2012

@figgybit I go into this in a bit more detail in the blog post, but I think the main difference between your first and second examples is that in the first, the exec'd code object is operating on locals, while in the second there are no locals (in module scope, locals and globals are the same; that is, there are only globals). Thus, in the second case, after execing the code object, the variable x no longer exists in the (global) scope; in the first example, the variable x is deleted in the exec'd code object's frame's locals, but not in the wrapping function's frame's locals.

@varun06
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varun06 commented May 1, 2012 via email

@figgybit
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figgybit commented May 1, 2012

@dcrosta thanks for the puzzle it was fun and I enjoy your explanation. Tell Mr. Diamond I say 'HI'

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