Willem Broekema wrote:
The issue came up while trying to get some Sympy code running on CLPython.
class C:
exec "a = 3"
print locals()
1. Is it guaranteed that class C gets an attribute "a", i.e. that the
locals printed include {'a': 3}?
2. It it (also) guaranteed if it were in a function scope?
The complete syntax of exec is:
exec CODE in Y, Z
where Y, Z are optional.
The documentation of "exec" says "if the optional parts are
omitted,the code is executed in the current scope." There are at least
two different interpretations:
a. The code is executed in the current class scope, so the assignment
must have an effect on the class scope.
b. The scope defaults to the local scope, by which is meant the
mapping returned by locals(), and of locals() the documentation says
that changes made to it may not influence the interpreter. (The
documentation of exec suggests using globals() and locals() as
arguments to exec, which seems hint at this interpretation.)
The relevant documentation:
exec:
http://docs.python.org/reference/simple_stmts.html#grammar-token-exec_stmt
locals: http://docs.python.org/library/functions.html#locals
The 3.0 doc for exec() has this warning:
"Warning
The default locals act as described for function locals() below:
modifications to the default locals dictionary should not be attempted.
Pass an explicit locals dictionary if you need to see effects of the
code on locals after function exec() returns."
This implies interpretation b.
However, is spite of the warning, class locals is a dict and locals() is
that dict, so a is available for further use in class code.
So the answer to question 1 for current CPython is yes.
Whether that is guaranteed for all implementations and versions is
another story.
Functions are much trickier. The local namespace is not a dict, and
modifying the locals() dict does not modify the namespace. The answer
to 2. is No, not even now.
>>> def f():
exec('a=3')
print(locals())
return a
>>> f()
{'a': 3}
Traceback (most recent call last):
File "<pyshell#37>", line 1, in <module>
f()
File "<pyshell#36>", line 4, in f
return a
NameError: global name 'a' is not defined
But why then is 'a' printed in the second call to locals (the implied
one in exec being the first)? It appears that a function or code object
can have only only one repeatedly used shadow dict. The 3.0 (and 2.5)
doc says "locals()
Update and return a dictionary representing the current local symbol
table." Note "update"; I had missed that before. To see this...
>>> def g():
a = locals()
b = locals()
return id(a), id(b), a,b
>>> g()
(20622048, 20622048, {'a': {...}}, {'a': {...}})
Inserting "print(a['a'])" between the locals calls raises KeyError.
Terry Jan Reedy
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