On Fri, Nov 20, 2015 at 11:39 PM, BartC <b...@freeuk.com> wrote:
> * The refusal to acknowledge that the def fn(a=[]) syntax is misleading.
> (What value will a have when you call fn()? The true answer is that you
> can't tell.)
It isn't misleading. The default value for the argument is set when
the function is defined, and it is set to the *object* that results
from evaluating the *expression* given. After that, the *object* is
the one you will always get (barring shenanigans) if the argument is
omitted. If the value of that object changes, it changes!
You keep expecting the *value* to be consistent. But what you actually
get is that the *object* is consistent. It's virtually impossible to
guarantee the former, in the face of mutable objects.
> * The persistent nonsense that somehow [] is mutable (what happens is that
> [] is assigned to a variable, and /that/ is mutable) (And I will probably
> get some flak now because 'assign' and 'variable' are meaningless in
> Python!)
What happens is that [] evaluates to an object, and *that object* is
mutable. Python does not store the syntactic element "[]" (empty
list-display), but instead stores (a reference to) the object. It's
exactly the same as:
def fn(a=list()):
...
def generate_list():
return []
def fn(a=generate_list()):
...
initial_list = []
def fn(a=initial_list):
...
import sys
def fn(a=[x*3+424 for x in range(not sys)]):
"""how ridiculous would you like to go?"""
Every one of these constructs exactly one new list, *when it is
evaluated*, which is at function definition time.
ChrisA
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