How to decide if a object is instancemethod?

[Cosmia Luna]

class Foo(object):
def bar(self):
return 'Something'

func = Foo().bar

if type(func) == <type 'instancemethod'>: # This should be always true
pass # do something here

What should type at <type 'instancemethod'>?

Thanks
Cosmia

 

[Jon Clements]

class Foo(object):
def bar(self):
return 'Something'

func = Foo().bar

if type(func) == <type 'instancemethod'>: # This should be always true
pass # do something here

What should type at <type 'instancemethod'>?

Thanks
Cosmia

import inspect
if inspect.ismethod(foo):
# ...

Will return True if foo is a bound method.

hth

Jon

 

[Steven D'Aprano]

But under what other conditions will it return True? The name suggests
that *any* method – static method, class method, bound method, unbound
method – will also result in True.

The documentation says only “instance methodâ€, though. Confusing :-(

Bound and unbound methods are instance methods. To be precise, the
"method" in "(un)bound method" stands for instance method, and the
difference between the two in Python 2.x is a flag on the method object.
(Unbound methods are gone in Python 3.)

Class and static methods are not instance methods. I suppose it is
conceivable that you could have an unbound class method in theory, but I
can't see any way to actually get one in practice.

In Python, and probably most languages, a bare, unadorned "method" is
implied to be an instance method; "instance method" is (possibly) a
retronym to distinguish them from other, newer(?), types of method.

http://en.wikipedia.org/wiki/Retronym

 

[Jean-Michel Pichavant]

import inspect
if inspect.ismethod(foo):
# ...

Will return True if foo is a bound method.

hth

Jon

another alternative :

import types

if type(func) == types.MethodType:
pass

or possibly better

if isinstance(func, types.MethodType):
pass


JM