Bugs item #1387650, was opened at 2005-12-22 00:04
Message generated for change (Comment added) made by sambayer
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Category: Python Interpreter Core
Group: Python 2.3
Status: Open
Resolution: None
Priority: 5
Submitted By: Samuel Bayer (sambayer)
Assigned to: Nobody/Anonymous (nobody)
Summary: weird behavior when assigning locals() to a variable
Initial Comment:
Tried this on Python 2.3.3, under Redhat Linux
Enterprise 3, and on Python 2.3.5, under MacOS X
10.4.3. Don't have access to 2.4 anywhere, so I can't
test it, but I took a look at the bug queue and can't
find any reference to this problem.
The following function yields a KeyError when run:
def foo():
b = locals()
c = 5
print b["c"]
The following function does not:
def foo():
b = locals()
c = 5
locals()
print b["c"]
There's no typo there. After referencing locals()
again, without updating the value of b, the printout works.
Note that b and locals(), as I believe is intended, are
identical:
def foo():
b = locals()
c = 5
print id(b)
print b.has_key()
print id(locals())
print b.has_key()
yields, when run:
>>> foo()
285984
False
285984
True
This has GOT to be a bug.
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>Comment By: Samuel Bayer (sambayer)
Date: 2005-12-22 00:25
Message:
Logged In: YES
user_id=40146
OK, it doesn't got to be a bug. After staring at the
documentation for locals(), I realize that locals() is only
updated when it's called again.
So it's not a bug, but it is a little odd. It's the only
namespace-like object you can't reliably use as a format
string argument. Consider the following:
>>> b = globals()
>>> c = 5
>>> print "%(c)d" % b
This prints the expected value. Ditto this:
>>> Class Foo: pass
>>> a = Foo(); b = a.__dict__
>>> a.c = 5
>>> print "%(c)d" % b
Only with locals(), inside a function, does this pattern
fail. I imagine it would be expensive to make work, though.
Never mind...
----------------------------------------------------------------------
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https://sourceforge.net/tracker/?func=detail&atid=105470&aid=1387650&group_id=5470
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