我找到了一种方法，不用子类化tuple, namedtuple等。你所需要做的就是在初始化后禁用setattr和delattr(如果你想让一个集合成为不可变的，也要禁用setitem和delitem):

def __init__(self, *args, **kwargs):
    # something here

    self.lock()

其中lock可以是这样的:

@classmethod
def lock(cls):
    def raiser(*a):
        raise TypeError('this instance is immutable')

    cls.__setattr__ = raiser
    cls.__delattr__ = raiser
    if hasattr(cls, '__setitem__'):
        cls.__setitem__ = raiser
        cls.__delitem__ = raiser

你可以用这个方法创建类Immutable，并像我展示的那样使用它。

如果你不想在每个init中都写self.lock()，你可以用元类自动实现:

class ImmutableType(type):
    @classmethod
    def change_init(mcs, original_init_method):
        def __new_init__(self, *args, **kwargs):
            if callable(original_init_method):
                original_init_method(self, *args, **kwargs)

            cls = self.__class__

            def raiser(*a):
                raise TypeError('this instance is immutable')

            cls.__setattr__ = raiser
            cls.__delattr__ = raiser
            if hasattr(cls, '__setitem__'):
                cls.__setitem__ = raiser
                cls.__delitem__ = raiser

        return __new_init__

    def __new__(mcs, name, parents, kwargs):
        kwargs['__init__'] = mcs.change_init(kwargs.get('__init__'))
        return type.__new__(mcs, name, parents, kwargs)


class Immutable(metaclass=ImmutableType):
    pass

Test

class SomeImmutableClass(Immutable):
    def __init__(self, some_value: int):
        self.important_attr = some_value

    def some_method(self):
        return 2 * self.important_attr


ins = SomeImmutableClass(3)
print(ins.some_method())  # 6
ins.important_attr += 1  # TypeError
ins.another_attr = 2  # TypeError

我通过重写__setattr__创建了不可变类，并且如果调用者是__init__，则允许该集合:

import inspect
class Immutable(object):
    def __setattr__(self, name, value):
        if inspect.stack()[2][3] != "__init__":
            raise Exception("Can't mutate an Immutable: self.%s = %r" % (name, value))
        object.__setattr__(self, name, value)

这还不够，因为它允许任何人的___init__来改变对象，但你懂的。

这里没有包括的是完全不可变性……不仅仅是父对象，还有所有的子对象。例如，元组/frozensets可能是不可变的，但它所属的对象可能不是。下面是一个小的(不完整的)版本，它在执行不变性方面做得很好:

# Initialize lists
a = [1,2,3]
b = [4,5,6]
c = [7,8,9]

l = [a,b]

# We can reassign in a list 
l[0] = c

# But not a tuple
t = (a,b)
#t[0] = c -> Throws exception
# But elements can be modified
t[0][1] = 4
t
([1, 4, 3], [4, 5, 6])
# Fix it back
t[0][1] = 2

li = ImmutableObject(l)
li
[[1, 2, 3], [4, 5, 6]]
# Can't assign
#li[0] = c will fail
# Can reference
li[0]
[1, 2, 3]
# But immutability conferred on returned object too
#li[0][1] = 4 will throw an exception

# Full solution should wrap all the comparison e.g. decorators.
# Also, you'd usually want to add a hash function, i didn't put
# an interface for that.

class ImmutableObject(object):
    def __init__(self, inobj):
        self._inited = False
        self._inobj = inobj
        self._inited = True

    def __repr__(self):
        return self._inobj.__repr__()

    def __str__(self):
        return self._inobj.__str__()

    def __getitem__(self, key):
        return ImmutableObject(self._inobj.__getitem__(key))

    def __iter__(self):
        return self._inobj.__iter__()

    def __setitem__(self, key, value):
        raise AttributeError, 'Object is read-only'

    def __getattr__(self, key):
        x = getattr(self._inobj, key)
        if callable(x):
              return x
        else:
              return ImmutableObject(x)

    def __hash__(self):
        return self._inobj.__hash__()

    def __eq__(self, second):
        return self._inobj.__eq__(second)

    def __setattr__(self, attr, value):
        if attr not in  ['_inobj', '_inited'] and self._inited == True:
            raise AttributeError, 'Object is read-only'
        object.__setattr__(self, attr, value)

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__()。

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

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)
>>>