[ python-Bugs-1387650 ] weird behavior when assigning locals() to a variable

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Bugs item #1387650, was opened at 2005-12-21 16:04
Message generated for change (Comment added) made by nnorwitz
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Category: Python Interpreter Core
Group: Python 2.3
>Status: Closed
>Resolution: Invalid
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: Neal Norwitz (nnorwitz)
Date: 2005-12-21 21:48

Message:
Logged In: YES 
user_id=33168

If you have suggestions for how we can improve the doc, let
us know.  I'll close this bug report since you seem to agree
it's not a bug.

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Comment By: Samuel Bayer (sambayer)
Date: 2005-12-21 16:25

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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