Re: storing references instead of copies in a dictionary

[mk]

def f2(arg):
    return "f2 "+arg
def f1(arg):
    return "f1 "+arg

a={"1":"f1","2":"f2"}
print [eval(x[1])(x[0]) for x in a.items()]
def f2(arg):
    return "New f2 "+arg
print [eval(x[1])(x[0]) for x in a.items()]

Neat trick, if probably dangerous in some circumstances. Anyway, thanks, 
I didn't think of that.

Don't know if this is any use to you..

At least I learned something. :-)




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