Consider: A)
.... pass
....
.... print args
....
<unbound method C.f>
(<__main__.C object at 0x04A51170>,)
And B)
<unbound method C.f>
(<__main__.C object at 0x04A51290>,)
I don't understand A). It is my vague understanding, that methods are
really properties that handle binding on attribute access, so B) should
be the "right" way to add a method to a class after definition. Why does
A show up as a method? Shouldn't it still just be a function? Certainly
when you define a class, there is some magic in the __new__ method that
turns functions in the initial dictionary into methods, but does this still
happen for all setattr after that? Is is possible to set a class attribute
equal to a regular (types.Function Type) function?
Any references that discuss these issues would be greatly appreciated.
Thanks,
Gerard
>>class C(object):
....
>>def f(*args):
....
>>C.f = f
>>C.f
>>C.f
>>c=C()
>>c.f()
>>c.f()
And B)
>>del c
>>C.f = types.MethodTyp e(f, None, C)
>>C.f
>>C.f = types.MethodTyp e(f, None, C)
>>C.f
>>c = C()
>>c.f()
>>c.f()
I don't understand A). It is my vague understanding, that methods are
really properties that handle binding on attribute access, so B) should
be the "right" way to add a method to a class after definition. Why does
A show up as a method? Shouldn't it still just be a function? Certainly
when you define a class, there is some magic in the __new__ method that
turns functions in the initial dictionary into methods, but does this still
happen for all setattr after that? Is is possible to set a class attribute
equal to a regular (types.Function Type) function?
Any references that discuss these issues would be greatly appreciated.
Thanks,
Gerard