On Tuesday, January 21, 2014 9:46:16 PM UTC+2, Chris Angelico wrote:
> On Wed, Jan 22, 2014 at 6:36 AM, Mû <m...@melix.net> wrote:
> > These were clear and quick answers to my problem. I did not think of this
> > possibility: the default argument is created once, but accessible only by
> > the function, therefore is not a global variable, whereas it looks like if
> > it were at first glance.
> You can actually poke at the function a bit and see what's happening.
> Try this in the interactive interpreter:
> >>> def f(x=[2,3]):
> x.append(1)
> return x
> >>> f()
> [2, 3, 1]
> >>> f()
> [2, 3, 1, 1]
> >>> f.__defaults__
> ([2, 3, 1, 1],)
>
> The __defaults__ attribute of a function is a tuple of its parameter
> defaults. You can easily see there that the list has changed as you
> changed it in the function. You could check it with id() or is, too:
> >>> id(f.__defaults__[0])
> 24529576
> >>> id(f())
> 24529576
> >>> f() is f.__defaults__[0]
> True
> ChrisA
that reminds me C's static :-)
def func(y, x = [1]):
if y != 1 :
func.__defaults__[0][0] = y
print(func.__defaults__[0])
func(0)
func(2)
func(1)
[0]
[2]
[2]
p.s. Mu, thanks for question!
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