I think it's wonderful that Python 3000 will have the nonlocal keyword
allowing complete lexical closures. I've just downloaded Python 3.0 a2 and
tried the following code:
def accumulator(n):
def foo(x):
nonlocal n
n = n + x
return n
return foo
... and it works!!! (Now let's see what Paul Graham has to say ... ;) )
However, I've also tried this:
def accumulator(n):
return lambda x: nonlocal n = n + x
and unfortunately it gave me a syntax error. I don't see any reason why this
kind of code should not be possible.
What do you all think?
--
Rocco Rossi
-----------------------------------------------------------------
"Alcuni vedono le cose come sono e dicono perché? Io sogno cose non ancora
esistite e chiedo perché no?"
G.B. Shaw
_______________________________________________
Python-3000 mailing list
[email protected]
http://mail.python.org/mailman/listinfo/python-3000
Unsubscribe:
http://mail.python.org/mailman/options/python-3000/archive%40mail-archive.com
