我使用了与Alex相同的想法:一个元类和一个“init marker”，但结合重写__setattr__:

>>> from abc import ABCMeta
>>> _INIT_MARKER = '_@_in_init_@_'
>>> class _ImmutableMeta(ABCMeta):
... 
...     """Meta class to construct Immutable."""
... 
...     def __call__(cls, *args, **kwds):
...         obj = cls.__new__(cls, *args, **kwds)
...         object.__setattr__(obj, _INIT_MARKER, True)
...         cls.__init__(obj, *args, **kwds)
...         object.__delattr__(obj, _INIT_MARKER)
...         return obj
...
>>> def _setattr(self, name, value):
...     if hasattr(self, _INIT_MARKER):
...         object.__setattr__(self, name, value)
...     else:
...         raise AttributeError("Instance of '%s' is immutable."
...                              % self.__class__.__name__)
...
>>> def _delattr(self, name):
...     raise AttributeError("Instance of '%s' is immutable."
...                          % self.__class__.__name__)
...
>>> _im_dict = {
...     '__doc__': "Mix-in class for immutable objects.",
...     '__copy__': lambda self: self,   # self is immutable, so just return it
...     '__setattr__': _setattr,
...     '__delattr__': _delattr}
...
>>> Immutable = _ImmutableMeta('Immutable', (), _im_dict)

注意:我直接调用元类，以使它在Python 2中都能工作。X和3.x。

>>> class T1(Immutable):
... 
...     def __init__(self, x=1, y=2):
...         self.x = x
...         self.y = y
...
>>> t1 = T1(y=8)
>>> t1.x, t1.y
(1, 8)
>>> t1.x = 7
AttributeError: Instance of 'T1' is immutable.

它也适用于插槽…:

>>> class T2(Immutable):
... 
...     __slots__ = 's1', 's2'
... 
...     def __init__(self, s1, s2):
...         self.s1 = s1
...         self.s2 = s2
...
>>> t2 = T2('abc', 'xyz')
>>> t2.s1, t2.s2
('abc', 'xyz')
>>> t2.s1 += 'd'
AttributeError: Instance of 'T2' is immutable.

．.． 和多重继承:

>>> class T3(T1, T2):
... 
...     def __init__(self, x, y, s1, s2):
...         T1.__init__(self, x, y)
...         T2.__init__(self, s1, s2)
...
>>> t3 = T3(12, 4, 'a', 'b')
>>> t3.x, t3.y, t3.s1, t3.s2
(12, 4, 'a', 'b')
>>> t3.y -= 3
AttributeError: Instance of 'T3' is immutable.

但是请注意，可变属性仍然是可变的:

>>> t3 = T3(12, [4, 7], 'a', 'b')
>>> t3.y.append(5)
>>> t3.y
[4, 7, 5]

最简单的方法是使用__slots__:

class A(object):
    __slots__ = []

A的实例现在是不可变的，因为您不能在它们上设置任何属性。

如果你想让类实例包含数据，你可以将this和derived from tuple结合起来:

from operator import itemgetter
class Point(tuple):
    __slots__ = []
    def __new__(cls, x, y):
        return tuple.__new__(cls, (x, y))
    x = property(itemgetter(0))
    y = property(itemgetter(1))

p = Point(2, 3)
p.x
# 2
p.y
# 3

编辑:如果你想摆脱索引，你可以重写__getitem__():

class Point(tuple):
    __slots__ = []
    def __new__(cls, x, y):
        return tuple.__new__(cls, (x, y))
    @property
    def x(self):
        return tuple.__getitem__(self, 0)
    @property
    def y(self):
        return tuple.__getitem__(self, 1)
    def __getitem__(self, item):
        raise TypeError

注意，不能使用operator。在这种情况下，属性的itemgetter，因为这将依赖于Point.__getitem__()而不是tuple.__getitem__()。此外，这不会阻止使用元组。__getitem__(p, 0)，但我很难想象这应该如何构成一个问题。

我不认为创建不可变对象的“正确”方法是编写C扩展。Python通常依赖于库实现者和库用户是成年人，而不是真正强制执行接口，接口应该在文档中清楚地说明。这就是为什么我不认为通过调用object.__setattr__()来规避被重写的__setattr__()是一个问题的可能性。如果有人这么做，风险自负。

除了其他优秀的答案之外，我喜欢为python 3.4(或者可能是3.3)添加一个方法。这个答案建立在之前对这个问题的几个答案的基础上。

在python 3.4中，可以使用不带设置符的属性来创建不可修改的类成员。(在早期版本中，可以不使用setter为属性赋值。)

class A:
    __slots__=['_A__a']
    def __init__(self, aValue):
      self.__a=aValue
    @property
    def a(self):
        return self.__a

你可以这样使用它:

instance=A("constant")
print (instance.a)

它会输出constant

而是调用实例。A =10会导致:

AttributeError: can't set attribute

解释:不带设置符的属性是python 3.4(我认为是3.3)的最新特性。如果您尝试给这样的属性赋值，则会引发Error。 使用插槽，我将成员变量限制为__A_a(即__a)。

问题:赋值给_aa仍然是可能的(instance. _aa =2)。但是如果你给一个私有变量赋值，那是你自己的错…

然而，这个答案不鼓励使用__slots__。使用其他方法来阻止属性创建可能更可取。

从Python 3.7开始，你可以在你的类中使用@dataclass装饰器，它将像结构体一样是不可变的!不过，它可能会也可能不会将__hash__()方法添加到类中。引用:

hash() is used by built-in hash(), and when objects are added to hashed collections such as dictionaries and sets. Having a hash() implies that instances of the class are immutable. Mutability is a complicated property that depends on the programmer’s intent, the existence and behavior of eq(), and the values of the eq and frozen flags in the dataclass() decorator. By default, dataclass() will not implicitly add a hash() method unless it is safe to do so. Neither will it add or change an existing explicitly defined hash() method. Setting the class attribute hash = None has a specific meaning to Python, as described in the hash() documentation. If hash() is not explicit defined, or if it is set to None, then dataclass() may add an implicit hash() method. Although not recommended, you can force dataclass() to create a hash() method with unsafe_hash=True. This might be the case if your class is logically immutable but can nonetheless be mutated. This is a specialized use case and should be considered carefully.

下面是上面链接的文档中的例子:

@dataclass
class InventoryItem:
    '''Class for keeping track of an item in inventory.'''
    name: str
    unit_price: float
    quantity_on_hand: int = 0

    def total_cost(self) -> float:
        return self.unit_price * self.quantity_on_hand

另一个想法是完全不允许__setattr__而使用object。构造函数中的__setattr__:

class Point(object):
    def __init__(self, x, y):
        object.__setattr__(self, "x", x)
        object.__setattr__(self, "y", y)
    def __setattr__(self, *args):
        raise TypeError
    def __delattr__(self, *args):
        raise TypeError

当然你可以用object。__setattr__(p， "x"， 3)来修改一个Point实例p，但您的原始实现遭受同样的问题(尝试tuple。__setattr__(i， "x"， 42)在一个不可变实例)。

您可以在原始实现中应用相同的技巧:去掉__getitem__()，并在属性函数中使用tuple.__getitem__()。

我不认为这是完全可能的，除非使用一个元组或namedtuple。无论如何，如果你重写了__setattr__()，用户总是可以通过直接调用object.__setattr__()来绕过它。任何依赖__setattr__的解决方案都保证不起作用。

以下是不使用某种元组可以得到的最接近的结果:

class Immutable:
    __slots__ = ['a', 'b']
    def __init__(self, a, b):
        object.__setattr__(self, 'a', a)
        object.__setattr__(self, 'b', b)
    def __setattr__(self, *ignored):
        raise NotImplementedError
    __delattr__ = __setattr__

但如果你足够努力，它就会破裂:

>>> t = Immutable(1, 2)
>>> t.a
1
>>> object.__setattr__(t, 'a', 2)
>>> t.a
2

但Sven对namedtuple的使用确实是不可变的。

更新

由于这个问题已经更新为询问如何在C中正确地做这件事，下面是我关于如何在Cython中正确地做这件事的答案:

第一个immutable.pyx:

cdef class Immutable:
    cdef object _a, _b

    def __init__(self, a, b):
        self._a = a
        self._b = b

    property a:
        def __get__(self):
            return self._a

    property b:
        def __get__(self):
            return self._b

    def __repr__(self):
        return "<Immutable {0}, {1}>".format(self.a, self.b)

和一个setup.py来编译它(使用命令setup.py build_ext——inplace:

from distutils.core import setup
from distutils.extension import Extension
from Cython.Distutils import build_ext

ext_modules = [Extension("immutable", ["immutable.pyx"])]

setup(
  name = 'Immutable object',
  cmdclass = {'build_ext': build_ext},
  ext_modules = ext_modules
)

然后试试吧:

>>> from immutable import Immutable
>>> p = Immutable(2, 3)
>>> p
<Immutable 2, 3>
>>> p.a = 1
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
AttributeError: attribute 'a' of 'immutable.Immutable' objects is not writable
>>> object.__setattr__(p, 'a', 1)
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
AttributeError: attribute 'a' of 'immutable.Immutable' objects is not writable
>>> p.a, p.b
(2, 3)
>>>