问题
If I have a
class A:
def foo(self):
pass
this evaluates to True:
getattr(A, 'foo') is A.foo
but this evaluates to False:
a = A()
getattr(a, 'foo') is a.foo
as does
a.foo is a.foo
Why?
I found that getattr(a, 'foo') and a.foo both are represented by
<bound method A.foo of <__main__.A object at 0x7a2c4de10d50>>)
So no hint there....
回答1:
At least in CPython, bound methods are implemented as an instance of a class method. Every time you ask for the value of a bound function, you get a new instance of this class.
x = a.foo
y = a.foo
assert x is not y
id(x) # E.g. 139664144271304
id(y) # E.g. 139664144033992
type(x) # <class 'method'>
type(y) # <class 'method'>
All this class does is store a reference to the instance and the unbound function, and when you call the class it calls the unbound function with the stored instance (along with your other arguments).
Unbound functions, like A.foo, are just regular old functions - no new instances of proxy classes are being constructed, so identity works as you expect.
The reason for this difference is that the semantic meaning of a.foo depends on two things, the value of a and the value of A.foo. In order to be able to get this meaning at any point in time later, both of these values need to be stored. This is what the method class does.
Conversely, the meaning of A.foo depends only on a single value: A.foo. So no additional work is required to store anything, and the value itself is used.
You might consider the idea of pre-allocating bound method instances, so that a.foo always returns the same immutable object - but given the dynamic nature of Python, it is simpler and cheaper to just construct a new one each time, even if they could be the same.
回答2:
To add to @GManNickG answer:
getattr(a, 'foo').__func__ is a.foo.__func__
will return True.
回答3:
Some objects stored in classes are descriptors, which don't follow normal rules for object lookups. The foo method you're dealing with in your example is one (function objects are descriptors).
A descriptor is an instance of a class that defines a __get__ (and optionally __set__ and __delete__) method(s). Those methods control what happens when you look up the desciptor on an instance of the class it's stored in.
I think an example will make this more clear:
class DescriptorClass:
def __get__(*args):
print("__get__ was called")
return "whatever"
class OtherClass:
descriptor_instance = DescriptorClass() # the descriptor instance is a class variable
other_instance = OtherClass()
# this calls the descriptor's __get__ method, which prints "__get__ was called"
result = other_instance.descriptor_instance
print(result) # will print "whatever", since that's what the __get__ method returned
A __get__ method doesn't need to return the same thing every time it's called. In fact, it usually won't. In the specific case of functions being used as descriptors (i.e. methods), a new "bound method" object will be created each time you look the function up. Thus the is operator will not see multiple bound methods as the same object, even though they may be binding the same function to the same instance.
来源：https://stackoverflow.com/questions/50498770/why-does-instance-of-object-foo-is-instance-of-object-foo-evaluate-false