-
-
Save dcrosta/2562350 to your computer and use it in GitHub Desktop.
| 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? | |
Thanks @figgybit. As a Python noob, I was assuming that was happening and now I know it was.
@varun06 I've posted my blog entry on this mystery: http://late.am/post/2012/04/30/the-exec-statement-and-a-python-mystery
@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.
@dcrosta thanks for the puzzle it was fun and I enjoy your explanation. Tell Mr. Diamond I say 'HI'
Hackers Unite
well this might enlighten the gist.
def test3():
x = 1
co2 = compile("""del x""", '', 'exec')
exec co2
print x
return x
This actually doesn't error but you would expect it to because the statement below fails.
x = 1
co2 = compile("""del x""", '', 'exec')
exec co2
print x
the reason is all about compiling. first we compile the codeobj (co2) and then the function is compiled. but when we compile the function the codeobj is not evalulated and as a result the deletion of the attribute x doesn't actually effect the results of the function.
so when we call the function it doesn't crash, YO!