Steve,
I don't think I understand. Here is what I just tried:
'>>> def f(): x = 3 d = locals() print x print d['x'] d['x'] = 5 print x
'>>> f()
3
3
3
'>>>
In your example, x had not yet been initialised, maybe. What I am seeing is that "x" does not seem to update when being assigned to - I guess this is what you are referring to by being unreliable.
Yes, that was my intent. In the same way that my "x" was not initialized, your "x" is not updated. locals() is readable but not writable in any case where locals() is not globals(), I believe.
But "unreliable" sounds kinda vague and intermittent, and I assume that is not the case here - What is the Rule(tm)?
Unfortunately, I don't think there is a Rule(tm) because the behavior of locals() (and globals()?) are implementation details. I remember hearing an explanation of why locals() is not writable that had to do with something about efficiency and the call stack (or something like that)...
Steve -- http://mail.python.org/mailman/listinfo/python-list
