"spir" <denis.s...@free.fr> wrote
x = 1
def f():
n = 1
def g0(a):
print (x + n + a)
return g0
I'm surprised the snippet below works as expected (py 2.6) without any
trick:
I'm not sure how else it could work.
x is a global name so the function must reference it.
n is a local name so it musdt evaluate it (it will disappear otherwise)
a is a parameter delivered at run time.
This means a (real) closure is built for g0, or what?
Thought I would need instead to use the old trick of pseudo
default-parameters:
def f():
n = 1
def g0(a, n=n, x=x):
print (x + n + a)
return g0
I did wonder if you would need n=n but I didn't think you would need x=x.
Its an interesting example and I confess I don't fully understand how
Python's
naming/reference rules are working here.
to let the inner func g0 "remember" outer values.
Why is this idiom used, then? Has something changed, or do I miss
a relevant point?
I thought you might need to do it if n had been a parameter of f()... but
having tried it no, it works as above.
I look forward to the explanation.
Alan G.
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