Re: How do I do this? (eval() on the left hand side)

[Caleb Hattingh]

Thx Peter
I verified this situation myself with a post from Steven Bethard earlier  
(trying to use "locals" within a function definition).

I am convinced now that locals() doesn't work as (I) expected.  Steven  
says there was some or other reason why locals() as used in this context  
is not writable - Do you know why this is?  I really do not like  
guidelines like "may not work", "is unreliable" and so on.  Perhaps this  
is a character flaw, but I really do like to know what works, when it  
works, and when it doesn't work.

In this scenario, we can see it doesn't work.  To my eyes, it doesn't work  
*in the way I expect* (which is highly subjective, no argument there).   
Would this be a situation where it would be nice to have an exception  
thrown if locals() is assigned to in a scope where it is not writable?

It would also be nice if globals and locals behaved the same, differing  
only in scope (which is what I expected originally anyway).  But we can't  
have everything, I guess :)

Caleb
On Wed, 08 Dec 2004 20:49:53 +0100, Peter Otten <[EMAIL PROTECTED]> wrote:
Caleb Hattingh wrote:
In what way is it  
unreliable?ÃÂÃÂIÃÂcan'tÃÂseemÃÂtoÃÂcreateÃÂaÃÂsituationÃÂwhere
the update through globals and locals doesn't  
work.ÃÂÃÂÃÂAreÃÂyouÃÂreferring
Updates to a locals() dictionary may not be reflected by the variables in
the local scope, e. g.:
def f():
...     locals()["a"] = 1
...     print a
...
f()
Traceback (most recent call last):
  File "<stdin>", line 1, in ?
  File "<stdin>", line 3, in f
NameError: global name 'a' is not defined
def f():
...     a = 42
...     locals()["a"] = 1
...     print a
...
f()
42
Updating globals() should be safe.
Peter
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