问题
Given a reference to a method, is there a way to check whether the method is bound to an object or not? Can you also access the instance that it's bound to?
回答1:
def isbound(method):
return method.im_self is not None
def instance(bounded_method):
return bounded_method.im_self
User-defined methods:
When a user-defined method object is created by retrieving a user-defined function object from a class, its
im_self
attribute isNone
and the method object is said to be unbound. When one is created by retrieving a user-defined function object from a class via one of its instances, itsim_self
attribute is the instance, and the method object is said to be bound. In either case, the new method'sim_class
attribute is the class from which the retrieval takes place, and itsim_func
attribute is the original function object.
In Python 2.6 and 3.0:
Instance method objects have new attributes for the object and function comprising the method; the new synonym for
im_self
is__self__
, andim_func
is also available as__func__
. The old names are still supported in Python 2.6, but are gone in 3.0.
回答2:
In python 3 the __self__
attribute is only set on bound methods. It's not set to None
on plain functions (or unbound methods, which are just plain functions in python 3).
Use something like this:
def is_bound(m):
return hasattr(m, '__self__')
回答3:
The chosen answer is valid in almost all cases. However when checking if a method is bound in a decorator using chosen answer, the check will fail. Consider this example decorator and method:
def my_decorator(*decorator_args, **decorator_kwargs):
def decorate(f):
print(hasattr(f, '__self__'))
@wraps(f)
def wrap(*args, **kwargs):
return f(*args, **kwargs)
return wrap
return decorate
class test_class(object):
@my_decorator()
def test_method(self, *some_params):
pass
The print
statement in decorator will print False
.
In this case I can't find any other way but to check function parameters using their argument names and look for one named self
. This is also not guarantied to work flawlessly because the first argument of a method is not forced to be named self
and can have any other name.
import inspect
def is_bounded(function):
params = inspect.signature(function).parameters
return params.get('self', None) is not None
回答4:
im_self attribute (only Python 2)
来源:https://stackoverflow.com/questions/53225/how-do-you-check-whether-a-python-method-is-bound-or-not