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Copy assignment for class member functions

I have a class A with some member functions that all do the same thing.
class A:
def a():
... boilerplate code ...
b = c = d = a
For debugging reasons, I would like to know the name of each member function at runtime. But since they all point to the same memory address, they will have the same __name__ attribute and I cannot figure out a way to distinguish between A.a and A.b just by looking at the object.
a = A.a
b = A.b
a.__name__ == b.__name__ # this is true
# how do I tell the difference between a and b?
Is there a way to achieve this without manually creating the functions b, c and d with the same boilerplate code?

No. Objects and names in Python live in separate spaces. There's only one function object there, and the function object doesn't know through what name it was conjured.
If you were a masochist, I suppose it would be possible to get a traceback and look at the line of code that called you, but that's just not practical.
You could do something like:
def reala(self,me=None):
pass
def a(self):
return reala('a')
def b(self):
return reala('b')
...

Order of classes matters when using variables from each other

I am new to Python and didn't find an answer to the following problem:
I have two classes and want them to use variables from each other. Is there a simple way to do this because if I do it like this class a does not know that class b exists.
class a:
y=1
print(b.x)
class b:
x=1
print(a.y)
And how do I use overwrite the variables, the following code does not work:
class a:
y=b.x
class b:
x=1

You are executing print as part of the class definition. It executes as soon as python sees that line of code, before it's read the part about class b.
Instead, use functions inside the classes to execute code after the classes have been defined:
class a:
y=1
def go():
print(b.x)
class b:
x=1
def go():
print(a.y)
a.go()
b.go()

As I said in a comment, your code isn't making effective use of classes. Here's what I think would be better approach that offers more flexibility in working around the circular reference issue.
First the class definitions (which follow the PEP 8 naming convention guidelines):
class A:
def __init__(self, value, linked_value=None):
self.y = value
if isinstance(linked_value, B):
self.linked_value = linked_value.x
def print_linked_value(self):
print(self.linked_value)
class B:
def __init__(self, value, linked_value=None):
self.x = value
if isinstance(linked_value, A):
self.linked_value = linked_value.y
def print_linked_value(self):
print(self.linked_value)
Definitions like that provide two ways to set up the circular references:
By creating them separately, then explicitly linking them:
# First create instances of each class.
a = A(1)
b = B(42)
# Then link them.
a.linked_value = b.x
b.linked_value = a.y
a.print_linked_value() # -> 42
b.print_linked_value() # -> 1
*OR* by creating the first one without a linked value and leaving only the second needing to be linked manually.
# First create instances of each class, but link the second to the first
# when it's created.
a = A(1)
b = B(42, a) # Create and link to first.
# Then link the first to the second to complete the circular references.
a.linked_value = b.x
# Same result.
a.print_linked_value() # -> 42
b.print_linked_value() # -> 1
Final note: Another, more advanced alternative that can also be applied in situations like this by using the built-in property() function as a decorator to create "descriptors". Here's an answer to a somewhat related question that illustrating its use.

class A:
y = 1
def foo(self):
print B.x
class B:
x = 1
def bar(self):
print A.y
>>> A().foo()
2
>>> B().bar()
1
Use 'print' in some function definition.

class initialization in Python

I found that some classes contain a __init__ function, and some don’t. I’m confused about something described below.
What is the difference between these two pieces of code:
class Test1(object):
i = 1
and
class Test2(object):
def __init__(self):
self.i = 1
I know that the result or any instance created by these two class and the way of getting their instance variable are pretty much the same. But is there any kind of “default” or “hidden” initialization mechanism of Python behind the scene when we don’t define the __init__ function for a class? And why I can’t write the first code in this way:
class Test1(object):
self.i = 1
That’s my questions. Thank you very much!
Thank you very much Antti Haapala! Your answer gives me further understanding of my questions. Now, I understand that they are different in a way that one is a "class variable", and the other is a "instance variable". But, as I tried it further, I got yet another confusing problem.
Here is what it is. I created 2 new classes for understanding what you said:
class Test3(object):
class_variable = [1]
def __init__(self):
self.instance_variable = [2]
class Test4(object):
class_variable = 1
def __init__(self):
self.instance_variable = 2
As you said in the answer to my first questions, I understand the class_variable is a "class variable" general to the class, and should be passed or changed by reference to the same location in the memory. And the instance_variable would be created distinctly for different instances.
But as I tried out, what you said is true for the Test3's instances, they all share the same memory. If I change it in one instance, its value changes wherever I call it.
But that's not true for instances of Test4. Shouldn't the int in the Test4 class also be changed by reference?
i1 = Test3()
i2 = Test3()
>>> i1.i.append(2)
>>> i2.i
[1, 2]
j1 = Test4()
j2 = Test4()
>>> j1.i = 3
>>> j2.i
1
Why is that? Does that "=" create an "instance variable" named "i" without changing the original "Test4.i" by default? Yet the "append" method just handles the "class variable"?
Again, thank you for your exhaustive explanation of the most boring basic concepts to a newbie of Python. I really appreciate that!

In python the instance attributes (such as self.i) are stored in the instance dictionary (i.__dict__). All the variable declarations in the class body are stored as attributes of the class.
Thus
class Test(object):
i = 1
is equivalent to
class Test(object):
pass
Test.i = 1
If no __init__ method is defined, the newly created instance usually starts with an empty instance dictionary, meaning that none of the properties are defined.
Now, when Python does the get attribute (as in print(instance.i) operation, it first looks for the attribute named i that is set on the instance). If that fails, the i attribute is looked up on type(i) instead (that is, the class attribute i).
So you can do things like:
class Test:
i = 1
t = Test()
print(t.i) # prints 1
t.i += 1
print(t.i) # prints 2
but what this actually does is:
>>> class Test(object):
... i = 1
...
>>> t = Test()
>>> t.__dict__
{}
>>> t.i += 1
>>> t.__dict__
{'i': 2}
There is no i attribute on the newly created t at all! Thus in t.i += 1 the .i was looked up in the Test class for reading, but the new value was set into the t.
If you use __init__:
>>> class Test2(object):
... def __init__(self):
... self.i = 1
...
>>> t2 = Test2()
>>> t2.__dict__
{'i': 1}
The newly created instance t2 will already have the attribute set.
Now in the case of immutable value such as int there is not that much difference. But suppose that you used a list:
class ClassHavingAList():
the_list = []
vs
class InstanceHavingAList()
def __init__(self):
self.the_list = []
Now, if you create 2 instances of both:
>>> c1 = ClassHavingAList()
>>> c2 = ClassHavingAList()
>>> i1 = InstanceHavingAList()
>>> i2 = InstanceHavingAList()
>>> c1.the_list is c2.the_list
True
>>> i1.the_list is i2.the_list
False
>>> c1.the_list.append(42)
>>> c2.the_list
[42]
c1.the_list and c2.the_list refer to the exactly same list object in memory, whereas i1.the_list and i2.the_list are distinct. Modifying the c1.the_list looks as if the c2.the_list also changes.
This is because the attribute itself is not set, it is just read. The c1.the_list.append(42) is identical in behaviour to
getattr(c1, 'the_list').append(42)
That is, it only tries read the value of attribute the_list on c1, and if not found there, then look it up in the superclass. The append does not change the attribute, it just changes the value that the attribute points to.
Now if you were to write an example that superficially looks the same:
c1.the_list += [ 42 ]
It would work identical to
original = getattr(c1, 'the_list')
new_value = original + [ 42 ]
setattr(c1, 'the_list', new_value)
And do a completely different thing: first of all the original + [ 42 ] would create a new list object. Then the attribute the_list would be created in c1, and set to point to this new list. That is, in case of instance.attribute, if the attribute is "read from", it can be looked up in the class (or superclass) if not set in the instance, but if it is written to, as in instance.attribute = something, it will always be set on the instance.
As for this:
class Test1(object):
self.i = 1
Such thing does not work in Python, because there is no self defined when the class body (that is all lines of code within the class) is executed - actually, the class is created only after all the code in the class body has been executed. The class body is just like any other piece of code, only the defs and variable assignments will create methods and attributes on the class instead of setting global variables.

I understood my newly added question. Thanks to Antti Haapala.
Now, when Python does the get attribute (as in print(instance.i) operation, it first looks for the attribute named i that is set on the instance). If that fails, the i attribute is looked up on type(i) instead (that is, the class attribute i).
I'm clear about why is:
j1 = Test4()
j2 = Test4()
>>> j1.i = 3
>>> j2.i
1
after few tests. The code
j1.3 = 3
actually creates a new instance variable for j1 without changing the class variable. That's the difference between "=" and methods like "append".
I'm a newbie of Python coming from c++. So, at the first glance, that's weird to me, since I never thought of creating a new instance variable which is not created in the class just using the "=". It's really a big difference between c++ and Python.
Now I got it, thank you all.

Python reverse introspection

Lets suppose this example: Two siblings classes where one loads the other class as a new attribute and then i wish to use this attribute from the main class inside the sibling.
a = 2
class AN(object):
def __init__(self,a):
self.aplus = a + 2
self.BECls = BE(a)
class BE(object):
def __init__(self,a):
print a
def get_aplus(self):
????
c = AN(a)
and i'd like to do:
c.BECls.get_aplus()
and this shall return something like self.self.aplus (metaphorically), that would be 4
Resuming: get aplus attribute from AN inside BE class, without declaring as arguments, but doing a "Reverse introspection", if it possible, considering the 'a' variable must be already loaded trough AN.
Sorry if I not made myself clear but I've tried to simplify what is happening with my real code.
I guess the problem may be the technique i'm using on the classes. But not sure what or how make it better.
Thanks

OP's question:
get aplus attribute from AN inside BE class, without declaring as
arguments, but doing a "Reverse introspection", if it possible,
considering the 'a' variable must be already loaded trough AN.
The closest thing we have to "reverse introspection" is a search through gc.getreferrers().
That said, it would be better to simply make the relationship explicit
class AN(object):
def __init__(self,a):
self.aplus = a + 2
self.BECls = BE(self, a)
class BE(object):
def __init__(self, an_obj, a):
self.an_obj = an_obj
print a
def get_aplus(self):
return self.an_obj.aplus
if __name__ == '__main__':
a = 2
c = AN(a)
print c.BECls.get_aplus() # this returns 4

python self-less

this works in the desired way:
class d:
def __init__(self,arg):
self.a = arg
def p(self):
print "a= ",self.a
x = d(1)
y = d(2)
x.p()
y.p()
yielding
a= 1
a= 2
i've tried eliminating the "self"s and using a global statement in __init__
class d:
def __init__(self,arg):
global a
a = arg
def p(self):
print "a= ",a
x = d(1)
y = d(2)
x.p()
y.p()
yielding, undesirably:
a= 2
a= 2
is there a way to write it without having to use "self"?

"self" is the way how Python works. So the answer is: No! If you want to cut hair: You don't have to use "self". Any other name will do also. ;-)

Python methods are just functions that are bound to the class or instance of a class. The only difference is that a method (aka bound function) expects the instance object as the first argument. Additionally when you invoke a method from an instance, it automatically passes the instance as the first argument. So by defining self in a method, you're telling it the namespace to work with.
This way when you specify self.a the method knows you're modifying the instance variable a that is part of the instance namespace.
Python scoping works from the inside out, so each function (or method) has its own namespace. If you create a variable a locally from within the method p (these names suck BTW), it is distinct from that of self.a. Example using your code:
class d:
def __init__(self,arg):
self.a = arg
def p(self):
a = self.a - 99
print "my a= ", a
print "instance a= ",self.a
x = d(1)
y = d(2)
x.p()
y.p()
Which yields:
my a= -98
instance a= 1
my a= -97
instance a= 2
Lastly, you don't have to call the first variable self. You could call it whatever you want, although you really shouldn't. It's convention to define and reference self from within methods, so if you care at all about other people reading your code without wanting to kill you, stick to the convention!
Further reading:
Python Classes tutorial

When you remove the self's, you end up having only one variable called a that will be shared not only amongst all your d objects but also in your entire execution environment.
You can't just eliminate the self's for this reason.