What is difference between bound and unbound function in Python? Example will be better to understand.

Question

while len(word_letters) > 0 and lives > 0:
        # letters used
        # ' '.join(['a', 'b', 'cd']) --> 'a b cd'
        print('You have', lives, 'lives left and you have used these letters: ', ' '.join(used_letters))

        # what current word is (ie W - R D)
        word_list = [letters if letters in used_letters else '-' for letters in word]
        print(lives_visual_dict[lives])
        print('Current word: ', ' '.join(word_list))

https://www.youtube.com/watch?v=8ext9G7xspg
the part which i am finding difficult to understand appears at 32:15 of the above video.
i don't know how to highlight a part of a code while posting a question on this forum. this is my first post.
if someone can help than i would be very grateful.

Answer 1

Whenever you look up a method via class.name or instance.name, the method object is created a-new. Python uses the descriptor protocol to wrap the function in a method object each time.

So, when you look up id(C.foo), a new method object is created, you retrieve it's id (a memory address), then discard the method object again. Then you look up id(cobj.foo), a new method object created that re-uses the now freed memory address and you see the same value. The method is then, again, discarded (garbage collected as the reference count drops to 0).

Next, you stored a reference to the C.foo unbound method in a variable. Now the memory address is not freed (the reference count is 1, instead of 0), and you create a second method instance by looking up cobj.foo which has to use a new memory location. Thus you get two different values.

See the documentation for id():

Return the “identity” of an object. This is an integer (or long integer) which is guaranteed to be unique and constant for this object during its lifetime. Two objects with non-overlapping lifetimes may have the same id() value.

CPython implementation detail: This is the address of the object in memory.
Emphasis mine.

You can re-create a method using a direct reference to the function via the __dict__ attribute of the class, then calling the __get__ descriptor method:

>>> class C(object):
...     def foo(self):
...         pass
... 
>>> C.foo
<unbound method C.foo>
>>> C.__dict__['foo']
<function foo at 0x1088cc488>
>>> C.__dict__['foo'].__get__(None, C)
<unbound method C.foo>
>>> C.__dict__['foo'].__get__(C(), C)
<bound method C.foo of <__main__.C object at 0x1088d6f90>>
Note that in Python 3, the whole unbound / bound method distinction has been dropped; you get a function where before you'd get an unbound method, and a method otherwise, where a method is always bound:

>>> C.foo
<function C.foo at 0x10bc48dd0>
>>> C.foo.__get__(None, C)
<function C.foo at 0x10bc48dd0>
>>> C.foo.__get__(C(), C)
<bound method C.foo of <__main__.C object at 0x10bc65150>>