lambda closure question

[Carl Banks]

jfj wrote:
> Carl Banks wrote:
> > Ted Lilley wrote:
> >
> >
> >>Unfortunately, it doesn't work.  It seems the closure keeps track
of
> >>the variable fed to it dynamically - if the variable changes after
> >  [...]
> >>
> >>At least, that's the explanation I'm deducing from this behavior.
> >
> >
> > And that's the correct explanation, chief.
>
> >
> > It is intended that way.  As an example of why that is: consider a
> > nested function called "printvars()" that you could insert in
various
> > places within a function to print out the value of some local
variable.
> >  If you did that, you wouldn't want printvars to print the values
at
> > the time it was bound, would you?
>
> Allow me to disagree (and start a new "confused with closures"
thread:)
>
> We know that python does not have references to variables. To some
> newcomers this may seem annoying but eventually they understand the
> pythonic way and they do without them. But in the case of closures
> python supports references!

That's right.  It's called Practicality beats purity.

The question is what's more important: to be able to use a nested
function to reference the outer scope, or to be able to use a nested
function as a closure inside the function it was defined in?  (Both
approaches effectively allow using a nested function as a closure if
it's used after the function returns.)

The Gods decided, for practical reasons, the former was more important.
 My experience bears this out: I find that about half the nested
functions I use are to reference an outer scope, half I return as a
closure.  Only once or twice did I try to use a nested function inside
the defining function.  I would guess this is more or less typical of
how nested functions are used.  If so, it was the right decision.


-- 
CARL BANKS