CODEWITHSUNDEEP
pythonclassmethods
Erxin
Erxin

Reputation: 1856

Why the python class method is different with it's object method but both of them have the same id?

While I'm studying the python attributes and methods, and read to the Example 1.2. A function is more. I try to do a simple test base on the example, but the result makes me a little confused. 

>>> class foo(object):
...     def fun(self):
...         pass
...     
>>> f = foo()
>>> f.fun is foo.fun
False
>>> id(f.fun) 
36093064
>>> id(foo.fun)
36093064
>>>

python version, 2.7.5 operation sys, win8

Why the 'is' test return False but the id return the same value? I expected to see the difference ids in the beginning. Thanks in advance!

Upvotes: 1

Views: 126

Answers (1)

Lennart Regebro
Lennart Regebro

Reputation: 172229

What the id is is an internal implementation detail. The id is of little use to use as a programmer. In CPython it tends to be the memory address of the object, the object in this case being a function object.

In this case the object is the same, it's a function. In one case the function is bound to an instance, in the other case it's not. This means that when you call it, in the bound case it will automatically get the instance as the first parameter, and in the second case not.

But it is the same function.

So why does f.fun is foo.fun return False?

Because methods compare as not being the same. Even with themselves:

 >>> foo.fun is foo.fun
 False
 >>> f = foo()
 >>> f.fun is f.fun
 False

This is because these objects are wrappers. If you want to compare methods, compare the im_func attribute:

>>> bar = foo
>>> bar.fun.im_func is foo.fun.im_func
True
>>> f.fun.im_func is foo.fun.im_func
True

In Python 3, the unbound method is gone, and normal functions are used. So in Python 3 you will get:

>>> foo.fun is foo.fun
True

Bound methods still compare as false to themselves:

>>> f.fun is f.fun
False

And you need to compare using the __func__ attribute:

>>> f.fun.__func__ is foo.fun
True

Upvotes: 5

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