I am experiencing a strange behavior of a built-in function locals() in Python. It is hard to explain exactly, but please take a look at a code:
def Main():
def F(l=locals()): print 'F', id(l), l
a= 100
F()
print '1', id(locals()), locals()
F()
In the local function F, I am assigning locals() into l as a default value for enclosure. Since locals() is a dict, its reference is copied to l. So the last three lines should have the same result.
However the result is like this:
F 139885919456064 {}
1 139885919456064 {'a': 100, 'F': <function F at 0x7f39ba8969b0>}
F 139885919456064 {'a': 100, 'F': <function F at 0x7f39ba8969b0>}
The three print statements are called at almost the same time, and id of locals() and l are the same, but the first l used in F does not have content.
I cannot understand why this happened. Can anyone explain this phenomenon? Or is this a known/unknown bug?
Many thanks!
If you read the docs for the locals function, you'll see
Update and return a dictionary representing the current local symbol table. Free variables are returned by locals() when it is called in function blocks, but not in class blocks.
locals() doesn't just return a dict of local variables; it also updates the dict to reflect current local variable values.
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