Fernando Perez schrieb:
> Hi all,
>
> consider the following small example:
>
> """
> Small test to try to understand a strange subtlety with closures
> """
>
> def outer(nmax):
>
> aa = []
> for n in range(nmax):
>
> def a(y):
> return (y,n)
> print 'Closure and cell id:',id(a.func_closure),\
> id(a.func_closure[0])
> aa.append(a)
>
> return aa
>
> print 'Closure creation.'
> nmax = 3
> aa = outer(nmax)
>
> print
> print 'Closure use.'
> for n in range(nmax):
> print '%s:%s' % (n,aa[n]('hello'))
>
> ################## EOF #################
>
>
> If I run this, I get:
>
> planck[test]> python debug_closures.py
> Closure creation.
> Closure and cell id: 1075998828 1075618940
> Closure and cell id: 1075999052 1075618940
> Closure and cell id: 1075999084 1075618940
>
> Closure use.
> 0:('hello', 2)
> 1:('hello', 2)
> 2:('hello', 2)
>
>
> My confusion arises from the printout after 'closure use'. I was expecting
> that
> each new function 'a' created inside the loop in 'outer' would capture the
> value of n, therefore my expectation was to see a printout like:
>
> 0:('hello', 0)
> 1:('hello', 1)... etc.
>
> However, what happens is a bit different. As can be seen from the printouts
> of 'Closure and cell id', in each pass of the loop a new closure is created,
> but it reuses the *same* cell object every time. For this reason, all the
> closures end up sharing the scope with the values determined by the *last*
> iteration of the loop.
>
> This struck me as counterintuitive, but I couldn't find anything in the
> official docs indicating what the expected behavior should be. Any
> feedback/enlightenment would be welcome. This problem appeared deep inside a
> complicated code and it took me almost two days to track down what was going
> on...
It's a FAQ. The reason is that the created closures don't capture the
_value_, but the _name_. Plus of course the locals()-dictionary outside
the function a to perform the lookup of that name. Which has the value
bound to it in the last iteration.
Common cure for this is to create an a-local name that shadows the outer
variable and is simultaneously bound to the desired value:
def outer(nmax):
aa = []
for n in range(nmax):
foo = 'bar'
def a(y,n=n):
bar = foo
return (y,n)
print 'Closure and cell id:',id(a.func_closure),\
id(a.func_closure[0])
aa.append(a)
return aa
print 'Closure creation.'
nmax = 3
aa = outer(nmax)
print
print 'Closure use.'
for n in range(nmax):
print '%s:%s' % (n,aa[n]('hello'))
Notice the foo/bar - that was necessary to actually create a closure at
all (to keep your printing working), as python statically checks if
there needs one to be.
Diez
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