Why an object changes its "address" between adjacent calls?

[Chris Angelico]

On Sun, Jun 17, 2018 at 7:39 PM, sales at caprilion.com.tw
<sales at caprilion.com.tw> wrote:
> Jim Lee at 2018/6/17 PM 04:10 wrote:
>>
>>
>>
>> On 06/17/2018 12:08 AM, Jach Fong wrote:
>>>
>>> C:\Python34\Doc>py
>>> Python 3.4.4 (v3.4.4:737efcadf5a6, Dec 20 2015, 19:28:18) [MSC v.1600 32
>>> bit (Intel)] on win32
>>> Type "help", "copyright", "credits" or "license" for more information.
>>> >>> import tkinter as tk
>>> >>> root = tk.Tk()
>>> >>> tk.Label(root, text='label one', font='TkDefaultFont').pack()
>>> >>> from tkinter import font
>>> >>> font.nametofont('TkDefaultFont')
>>> <tkinter.font.Font object at 0x021E9490>
>>> >>> font.nametofont('TkDefaultFont')
>>> <tkinter.font.Font object at 0x021E9390>
>>> >>>
>>>
>>> The "address" of the Font object 'TkDefaultFont' changes, why?
>>>
>>> Best Regards,
>>> Jach Fong
>>>
>>>
>>
>> def nametofont(name):
>>      """Given the name of a tk named font, returns a Font representation.
>>      """
>>      return Font(name=name, exists=True)
>>
>> Every time you call nametofont(), you're creating a new instance of the
>> Font class.
>
>
>     hmm... It means every time I set a widget's font to "TkDefaultFont", a
> new object was created. Why python do things this way? Can't it use
> this same object again and again?
>

That would imply keeping the object around until it's needed again.
Sometimes that's worth doing; other times it isn't, or isn't worth the
hassle. As soon as you stop using the previous one, Python is free to
dispose of it.

ChrisA