On 2017-08-05, Tim Daneliuk wrote:
> On 08/05/2017 03:21 PM, Chris Angelico wrote:
>> so the object's lifetime shouldn't matter to you.
>
> I disagree with this most strongly. That's only true when the
> machine resources being consumed by your Python object are small in
>
On 08/05/2017 05:36 PM, Ned Batchelder wrote:
> On 8/5/17 5:41 PM, Tim Daneliuk wrote:
>> On 08/05/2017 11:16 AM, Ned Batchelder wrote:
>>> It uses
>>> reference counting, so most objects are reclaimed immediately when their
>>> reference count goes to zero, such as at the end of local scopes.
>>
On 08/05/2017 05:36 PM, Ned Batchelder wrote:
> On 8/5/17 5:41 PM, Tim Daneliuk wrote:
>> On 08/05/2017 11:16 AM, Ned Batchelder wrote:
>>> It uses
>>> reference counting, so most objects are reclaimed immediately when their
>>> reference count goes to zero, such as at the end of local scopes.
>>
On 08/05/2017 05:58 PM, Chris Angelico wrote:
> On Sun, Aug 6, 2017 at 7:32 AM, Tim Daneliuk wrote:
>> On 08/05/2017 03:21 PM, Chris Angelico wrote:
>>> After a 'with' block,
>>> the object *still exists*, but it has been "exited" in some way
>>> (usually by closing/releasing
On 08/05/2017 05:58 PM, Chris Angelico wrote:
> On Sun, Aug 6, 2017 at 7:32 AM, Tim Daneliuk wrote:
>> On 08/05/2017 03:21 PM, Chris Angelico wrote:
>>> After a 'with' block,
>>> the object *still exists*, but it has been "exited" in some way
>>> (usually by closing/releasing
On Sun, Aug 6, 2017 at 7:32 AM, Tim Daneliuk wrote:
> On 08/05/2017 03:21 PM, Chris Angelico wrote:
>> After a 'with' block,
>> the object *still exists*, but it has been "exited" in some way
>> (usually by closing/releasing an underlying resource).
>
> The containing object
On 8/5/17 5:41 PM, Tim Daneliuk wrote:
> On 08/05/2017 11:16 AM, Ned Batchelder wrote:
>> It uses
>> reference counting, so most objects are reclaimed immediately when their
>> reference count goes to zero, such as at the end of local scopes.
> Given this code:
>
> class SomeObject:
> .
>
On 2017-08-05 22:41, Tim Daneliuk wrote:
On 08/05/2017 11:16 AM, Ned Batchelder wrote:
It uses
reference counting, so most objects are reclaimed immediately when their
reference count goes to zero, such as at the end of local scopes.
Given this code:
class SomeObject:
.
for foo
Tim Daneliuk :
> Are you saying that each time a,b,c are reassigned to new instances of
> SomeObject the old instance counts go to 0 and are immediately - as in
> synchronously, right now, on the spot - removed from memory?
That depends on the implementation of Python.
On 08/05/2017 11:16 AM, Ned Batchelder wrote:
> It uses
> reference counting, so most objects are reclaimed immediately when their
> reference count goes to zero, such as at the end of local scopes.
Given this code:
class SomeObject:
.
for foo in somelist:
a = SomeObject(foo)
b
On 08/05/2017 03:21 PM, Chris Angelico wrote:
> After a 'with' block,
> the object *still exists*, but it has been "exited" in some way
> (usually by closing/releasing an underlying resource).
The containing object exists, but the things that the closing
logic explicitly released do not. In some
On Sun, Aug 6, 2017 at 1:23 AM, Tim Daneliuk wrote:
> On 08/04/2017 07:00 PM, Chris Angelico wrote:
>> Again, don't stress about exactly when objects get
>> disposed of; it doesn't matter.
>
>
> Respectfully, I disagree strongly. Objects get build on the heap and
> persist
On 8/5/17 11:23 AM, Tim Daneliuk wrote:
> On 08/04/2017 07:00 PM, Chris Angelico wrote:
>> Again, don't stress about exactly when objects get
>> disposed of; it doesn't matter.
>
> Respectfully, I disagree strongly. Objects get build on the heap and
> persist even when they go out of scope until
On 8/4/17 7:42 PM, Jon Forrest wrote:
> On 8/4/2017 4:34 PM, gst wrote:
>> 'two' is a so called constant or literal value .. (of that
>> function).
>>
>> Why not attach it, as a const value/object, to the function itself ?
>> So that a new string object has not to be created each time the
>>
Tim Daneliuk :
> On 08/04/2017 07:00 PM, Chris Angelico wrote:
>> Again, don't stress about exactly when objects get disposed of; it
>> doesn't matter.
>
> Respectfully, I disagree strongly. Objects get build on the heap and
> persist even when they go out of scope until such
On 08/04/2017 07:00 PM, Chris Angelico wrote:
> Again, don't stress about exactly when objects get
> disposed of; it doesn't matter.
Respectfully, I disagree strongly. Objects get build on the heap and
persist even when they go out of scope until such time garbage
collection takes place. This
On 8/4/2017 7:11 PM, Jon Forrest wrote:
Consider the following Python shell session (Python 3.6.2, Win64):
>>> def givemetwo():
... x = 'two'
... print(id(x))
...
>>> givemetwo()
1578505988392
So far fine. My understanding of object existence made me
think that the object
On Sat, 5 Aug 2017 09:11 am, Jon Forrest wrote:
> Consider the following Python shell session (Python 3.6.2, Win64):
>
> >>> def givemetwo():
> ... x = 'two'
> ... print(id(x))
> ...
> >>> givemetwo()
> 1578505988392
>
> So far fine. My understanding of object existence made
On Sat, Aug 5, 2017 at 9:47 AM, Jon Forrest wrote:
> Perhaps the reason the variable isn't destroyed is
> shown by the following (again, in the same session):
>
import sys
sys.getrefcount(1578505988392)
> 3
>
> So, maybe it's not destroyed because there are still
>
On Sat, Aug 5, 2017 at 9:42 AM, Jon Forrest wrote:
> On 8/4/2017 4:34 PM, gst wrote:
>>
>> 'two' is a so called constant or literal value .. (of that
>> function).
>>
>> Why not attach it, as a const value/object, to the function itself ?
>> So that a new string object has not
On 8/4/2017 4:34 PM, gst wrote:
'two' is a so called constant or literal value .. (of that
function).
Why not attach it, as a const value/object, to the function itself ?
So that a new string object has not to be created each time the
function is called. Because anyway strings are immutable. So
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