On 2015-11-19 00:34, fl wrote:
On Wednesday, November 18, 2015 at 7:15:05 PM UTC-5, Chris Angelico wrote:
On Thu, Nov 19, 2015 at 11:02 AM, Ian Kelly <ian.g.ke...@gmail.com> wrote:
> On Wed, Nov 18, 2015 at 4:22 PM, Chris Angelico <ros...@gmail.com> wrote:
>> On Thu, Nov 19, 2015 at 10:14 AM, BartC <b...@freeuk.com> wrote:
>>> So, looking at some source code, a default value for certain types is only
>>> certain to be that value for the very first call of that function?
>>
>> On the contrary, it is certain always to be that exact object.
>
> "Certain" may be a bit overly strong.
>
>>>> def f(x=42):
> ... return x
> ...
>>>> f()
> 42
>>>> f.__defaults__ = (43,)
>>>> f()
> 43
I'll raise you one.
>>> def f(x=42):
... return x
...
>>> f()
42
>>> import ctypes
>>> ctypes.c_int.from_address(id(43)+ ctypes.sizeof(ctypes.c_size_t) +
ctypes.sizeof(ctypes.c_voidp)).value=42
>>> f.__defaults__ = (43,)
>>> f()
42
>>>
Nothing is certain in Python. And two wrongs can make a... hmm... no,
this is not a right. It is not a privilege either. It is a dastardly
trick played on people's minds.
ChrisA
After I try with
list1 = eList(12, [2])
If you pass something into the parameter, the default isn't needed/used.
and
list1 = eList(12)
You're not passing anything into the parameter, so the default is used.
it gives me new surprises. Even though I delete list1, the subsequent
list1 = eList(12)
will remember the number of '12' of the previous sequence. This is my new
question: What does 'del' do? It does not look like a thorough list deletion
from the effect.
No, "del list1" doesn't delete the list, it deletes the name "list1".
The list still exists as the default parameter.
A default parameter is evaluated when the function is defined, and that
value exists as the default as long as the function exists.
If the value is mutable (can be modified), and you modify it, then the
function will see that modified value.
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