Re: storing references instead of copies in a dictionary

[mk]

Calvin Spealman wrote:

To your actual problem... Why do you wanna do this anyway? If you want
to change the function in the dictionary, why don't you simply define
the functions you'll want to use, and change the one you have bound to
the key in the dictionary when you want to change it? In other words,
define them all at once, and then just d['1'] = new_f1. What is wrong
with that?

Well, basically nothing except I need to remember I have to do that.

Suppose one does that frequently in a program. It becomes tedious. I
think I will define some helper function then:

>>def helper(fundict, newfun):

.... fundict[newfun.func_nam e] = newfun
....

_If_ there were some shorter and still "proper" way to do it, I'd use
it. If not, no big deal.

For completeness:

def new_f1(arg):
return "NEW f1 " + arg
f1.func_code = new_f1.func_cod e

Don't use that unless you really have to and I nearly promise that you don't.

I promise I won't use it. :-) It seems like a 'wrong thing to do'.

[castironpi]

On Jul 18, 1:32*am, John Machin <sjmac...@lexic on.netwrote:

On Jul 18, 4:26*pm, castironpi <castiro...@gma il.comwrote:

I delicately ask for an example in natural language and daily life in
which we change what object a name refers to,

her, him, it, ... i.e. any pronoun

In that case,

it= Dog( )
refs[ 'it' ]= it
it.walk( )
it= Cat( )
it.feed( )

you don't want refs[ 'it' ] to refer to cat. You were talking about
the dog when you stored refs[ 'it' ].

The OP might be wanting to mutate locals( ). If not define a
Reference object,

class Reference:
def __init__( self, ref ): self.ref= ref

a more descriptive name than Blank (above), to explicitly state that
you're using a reference instead of "fixed-bound" namespace semantics,
or whatever the right word is.