描述器类中__get__()的疑问

[explist]

import types

from functools import wraps

class profiled:

def __init__(self,func):
    wraps(func)(self)    
 
def __call__(self,*a,**k):
    print("profiled.call")        
    return self.__wrapped__(*a,**k)

def __get__(self,instance,cls):
    if instance is None: return self
    foo=types.MethodType(self,instance)
    print("profiled.get:",foo)
    return foo

class spam:

@profiled
def bar(self,x):print(x)

if name == 'main':

s=spam()
print(s.__dict__,spam.__dict__,'\n')
f=s.bar
print(f,type(f),f.__dict__,sep='\n')
f("TEST")  #f 已经是绑定的方法了，为什么还要调用__call__()

[explist]

输出：
{} {'__dict__': <attribute '__dict__' of 'spam' objects>, '__doc__': None, '__weakref__': <attribute '__weakref__' of 'spam' objects>, '__module__': '__main__', 'bar': <__main__.profiled object at 0x02A8BCF0>}

profiled.get: <bound method spam.bar of <__main__.spam object at 0x70>>
<bound method spam.bar of <__main__.spam object at 0x70>>
<class 'method'>
{'__annotations__': {}, '__wrapped__': <function spam.bar at 0x10>, '__qualname__': 'spam.bar', '__name__': 'bar', '__module__': '__main__', 'n': 0, '__doc__': None}
profiled.call
TEST

[billgreen1]

因为 f 不是 bar ，而是 wrappered 过的 bar,本质上应该是 profiled 类吧

[explist]

看了下 MethodType 的源码，不是很明白这里面的原理：

class _C:
def _m(self): pass
MethodType = type(_C()._m)

[explist]

@explist
type(_c()._m) #这是创建类还是查看类型？