When do default parameters get their values set?

[Chris Kaynor]

On Wed, Dec 10, 2014 at 7:15 PM, Steven D'Aprano <steve at pearwood.info>
wrote:

> On Wed, 10 Dec 2014 18:18:44 -0800, Rustom Mody wrote:
>
>
> > And going the other way -- no defs only lambdas its this:
> >
> >
> >>>> f = lambda : (lambda x= {}: x)
> >>>> f()() is f()()
> > False
> >>>> d = f()
> >>>> d() is d()
> > True
> >>>>
> >>>>
> >
> > But I have a different question -- can this be demonstrated without the
> > 'is'?
>
>
> Can *what* be demonstrated? That the functions returned are different
> objects, or that the dicts are different objects? Both? Something else?
>
> Using "is" you are demonstrating that calling the function twice returns
> two distinct objects. That is the purpose of "is", to compare object
> identity. Without "is", you can compare object IDs directly:
>
> id(f()()) == id(f()())
>
> but that's ugly and less efficient. Using "is" is the more idiomatic and
> natural way to do this.
>

In CPython, that does not work, as the dictionary will be garbage collected
after each call to id:
>>> f = lambda : (lambda x= {}: x)
>>> f()() is f()()
False
>>> id(f()()) == id(f()())
True


>
> Other than that, you could do something to demonstrate the consequences
> of the two values being distinct objects:
>
> a = f()()  # Call twice to get a dict.
> b = f()()
> a['key'] = 23
> b['key']  # raises KeyError
>
>
> or
>
> a = f()  # Call once to get a function.
> b = f()
> a.attribute = 23
> b.attribute  # raises AttributeError
>
>
> Or you could inspect the byte-code of f and try to understand it.
>
>
Another way would be to to demonstrate the behavior, without using is,
would be to use a global counter variable:

a = 1
def f():
    global a
    def g(b=[a]):
        return b
    a += 1
    return g

 >>> f()()
[1]
>>> f()()
[2]
>>> a = f()
>>> a()
[3]
>>> a()
[3]
>>> b = f()
>>> b()
[4]
>>> a()
[3]
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