Suppose we have the following code:
class T(object):
def m1(self, a):
...
f=T.m1
How to call f on an instance of T?
x=T()
x.f(..)?
f(x, ..)?
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2 Answers
A member function is just like any other function, except it takes self as the first argument and there is a mechanism which passes that argument automatically.
So, the short answer is, use it ths way:
class T(object):
def m1(self, a):
pass
f=T.m1
x = T()
f(x, 1234)
Unbound Method
This is because you are using T.m1, which is an "unbound method". Unbound here means that its self argument is not bound to an instance.
>>> T.m1
<unbound method T.m1>
Bound Method
Unlike T.m1, x.m1 gives you a bound method:
>>> x.m1
<bound method T.m1 of <__main__.T object at 0x0000000002483080>>
You can reference a bound method and use it without passing self explicitly:
f2 = x.m1
f2(1234)
Bind using partial
You can also do the equivalent "binding" self yourself, with this code:
import functools
unbound_f = T.m1
bound_f = functools.partial(unbound_f, x)
bound_f(1234)
2 Comments
RafazZ
You can also use the
@staticmethod decorator
zvone
@RafazZ If you used
@staticmethod you would have different behaviour. The function would not have a self argument. Here, m1 receives a self in each of the three cases I wrote. If self is not used within m1, then yes - @staticmethod would be the preferred solution.Hopefully this explains it!
class T(object):
def m1(self, a):
return a
f=T().m1 #must initialize the class using '()' before calling a method of that class
f(1)
f = T()
f.m1(1)
f = T().m1(1)
f = T
f().m1(1)
f = T.m1
f(T(), 1) #can call the method without first initializing the class because we pass a reference of the methods class
f = T.m1
f2 = T()
f(f2, 1)
