Pointers in Python?

There's no way you can do that changing only that line. You can do:

a = [1]
b = a
a[0] = 2
b[0]

That creates a list, assigns the reference to a, then b also, uses the a reference to set the first element to 2, then accesses using the b reference variable.

I want form.data['field'] and form.field.value to always have the same value

This is feasible, because it involves decorated names and indexing -- i.e., completely different constructs from the barenames a and b that you're asking about, and for with your request is utterly impossible. Why ask for something impossible and totally different from the (possible) thing you actually want?!

Maybe you don't realize how drastically different barenames and decorated names are. When you refer to a barename a, you're getting exactly the object a was last bound to in this scope (or an exception if it wasn't bound in this scope) -- this is such a deep and fundamental aspect of Python that it can't possibly be subverted. When you refer to a decorated name x.y, you're asking an object (the object x refers to) to please supply "the y attribute" -- and in response to that request, the object can perform totally arbitrary computations (and indexing is quite similar: it also allows arbitrary computations to be performed in response).

Now, your "actual desiderata" example is mysterious because in each case two levels of indexing or attribute-getting are involved, so the subtlety you crave could be introduced in many ways. What other attributes is form.field suppose to have, for example, besides value? Without that further .value computations, possibilities would include:

class Form(object):
   ...
   def __getattr__(self, name):
       return self.data[name]

and

class Form(object):
   ...
   @property
   def data(self):
       return self.__dict__

The presence of .value suggests picking the first form, plus a kind-of-useless wrapper:

class KouWrap(object):
   def __init__(self, value):
       self.value = value

class Form(object):
   ...
   def __getattr__(self, name):
       return KouWrap(self.data[name])

If assignments such form.field.value = 23 is also supposed to set the entry in form.data, then the wrapper must become more complex indeed, and not all that useless:

class MciWrap(object):
   def __init__(self, data, k):
       self._data = data
       self._k = k
   @property
   def value(self):
       return self._data[self._k]
   @value.setter
   def value(self, v)
       self._data[self._k] = v

class Form(object):
   ...
   def __getattr__(self, name):
       return MciWrap(self.data, name)

The latter example is roughly as close as it gets, in Python, to the sense of "a pointer" as you seem to want -- but it's crucial to understand that such subtleties can ever only work with indexing and/or decorated names, never with barenames as you originally asked!

It's not a bug, it's a feature :-)

When you look at the '=' operator in Python, don't think in terms of assignment. You don't assign things, you bind them. = is a binding operator.

So in your code, you are giving the value 1 a name: a. Then, you are giving the value in 'a' a name: b. Then you are binding the value 2 to the name 'a'. The value bound to b doesn't change in this operation.

Coming from C-like languages, this can be confusing, but once you become accustomed to it, you find that it helps you to read and reason about your code more clearly: the value which has the name 'b' will not change unless you explicitly change it. And if you do an 'import this', you'll find that the Zen of Python states that Explicit is better than implicit.

Note as well that functional languages such as Haskell also use this paradigm, with great value in terms of robustness.

Yes! there is a way to use a variable as a pointer in python!

I am sorry to say that many of answers were partially wrong. In principle every equal(=) assignation shares the memory address (check the id(obj) function), but in practice it is not such. There are variables whose equal("=") behaviour works in last term as a copy of memory space, mostly in simple objects (e.g. "int" object), and others in which not (e.g. "list","dict" objects).

Here is an example of pointer assignation

dict1 = {'first':'hello', 'second':'world'}
dict2 = dict1 # pointer assignation mechanism
dict2['first'] = 'bye'
dict1
>>> {'first':'bye', 'second':'world'}

Here is an example of copy assignation

a = 1
b = a # copy of memory mechanism. up to here id(a) == id(b)
b = 2 # new address generation. therefore without pointer behaviour
a
>>> 1

Pointer assignation is a pretty useful tool for aliasing without the waste of extra memory, in certain situations for performing comfy code,

class cls_X():
   ...
   def method_1():
      pd1 = self.obj_clsY.dict_vars_for_clsX['meth1'] # pointer dict 1: aliasing
      pd1['var4'] = self.method2(pd1['var1'], pd1['var2'], pd1['var3'])
   #enddef method_1
   ...
#endclass cls_X

but one have to be aware of this use in order to prevent code mistakes.

To conclude, by default some variables are barenames (simple objects like int, float, str,...), and some are pointers when assigned between them (e.g. dict1 = dict2). How to recognize them? just try this experiment with them. In IDEs with variable explorer panel usually appears to be the memory address ("@axbbbbbb...") in the definition of pointer-mechanism objects.

I suggest investigate in the topic. There are many people who know much more about this topic for sure. (see "ctypes" module). I hope it is helpful. Enjoy the good use of the objects! Regards, José Crespo

>> id(1)
1923344848  # identity of the location in memory where 1 is stored
>> id(1)
1923344848  # always the same
>> a = 1
>> b = a  # or equivalently b = 1, because 1 is immutable
>> id(a)
1923344848
>> id(b)  # equal to id(a)
1923344848

As you can see a and b are just two different names that reference to the same immutable object (int) 1. If later you write a = 2, you reassign the name a to a different object (int) 2, but the b continues referencing to 1:

>> id(2)
1923344880
>> a = 2
>> id(a)
1923344880  # equal to id(2)
>> b
1           # b hasn't changed
>> id(b)
1923344848  # equal to id(1)

What would happen if you had a mutable object instead, such as a list [1]?

>> id([1])
328817608
>> id([1])
328664968  # different from the previous id, because each time a new list is created
>> a = [1]
>> id(a)
328817800
>> id(a)
328817800 # now same as before
>> b = a
>> id(b)
328817800  # same as id(a)

Again, we are referencing to the same object (list) [1] by two different names a and b. However now we can mutate this list while it remains the same object, and a, b will both continue referencing to it

>> a[0] = 2
>> a
[2]
>> b
[2]
>> id(a)
328817800  # same as before
>> id(b)
328817800  # same as before