Message21612
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user_id=341410
>>> def f():
... y = 5
... print 'y' in globals(), 'y' in locals()
... def i():
... print 'y' in globals(), 'y' in locals()
... i()
...
>>> f()
False True
False False
>>>
>>> def g():
... gl = {};lo={}
... exec '''y = 5
... print 'y' in globals(), 'y' in locals()
... def i():
... print 'y' in globals(), 'y' in locals()
... i()
... ''' in gl, lo
...
>>> g()
False True
False False
That looks constant...but what if we print out 'y'?
>>> def s():
... y = 5
... print 'y' in globals(), 'y' in locals(), y
... def i():
... print 'y' in globals(), 'y' in locals(), y
... i()
...
>>> s()
False True 5
False True 5
>>>
>>> def t():
... gl = {};lo = {}
... exec '''y = 5
... print 'y' in globals(), 'y' in locals(), y
... def i():
... print 'y' in globals(), 'y' in locals(), y
... i()
... ''' in gl, lo
...
>>> t()
False True 5
False False
Traceback (most recent call last):
File "<stdin>", line 1, in ?
File "<stdin>", line 3, in t
File "<string>", line 5, in ?
File "<string>", line 4, in i
NameError: global name 'y' is not defined
Now why did 'y' stick itself into the locals() of i() in
s()? Is this another bug?
What if we remove the namespaces gl and lo?
>>> def u():
... exec '''y = 5
... print 'y' in globals(), 'y' in locals(), y
... def i():
... print 'y' in globals(), 'y' in locals(), y
... i()
... '''
...
>>> u()
False True 5
False False
Traceback (most recent call last):
File "<stdin>", line 1, in ?
File "<stdin>", line 2, in u
File "<string>", line 5, in ?
File "<string>", line 4, in i
NameError: global name 'y' is not defined
Nope, still dies. It does seem to be a bug in exec. I'm
still curious why 'y' was placed into i()'s local namespace
in s(). |
|
Date |
User |
Action |
Args |
2007-08-23 14:23:41 | admin | link | issue991196 messages |
2007-08-23 14:23:41 | admin | create | |
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