虽然我从来都不需要这样做,但我突然意识到用Python创建一个不可变对象可能有点棘手。你不能只是覆盖__setattr__,因为这样你甚至不能在__init__中设置属性。子类化一个元组是一个有效的技巧:

class Immutable(tuple):
    
    def __new__(cls, a, b):
        return tuple.__new__(cls, (a, b))

    @property
    def a(self):
        return self[0]
        
    @property
    def b(self):
        return self[1]

    def __str__(self):
        return "<Immutable {0}, {1}>".format(self.a, self.b)
    
    def __setattr__(self, *ignored):
        raise NotImplementedError

    def __delattr__(self, *ignored):
        raise NotImplementedError

但是你可以通过self[0]和self[1]访问a和b变量,这很烦人。

这在Pure Python中可行吗?如果不是,我该如何用C扩展来做呢?

(只能在python3中工作的答案是可以接受的)。

更新:

从Python 3.7开始,要使用的方法是使用@dataclass装饰器,参见最新接受的答案。


当前回答

这种方式不停止对象。__setattr__从工作,但我仍然发现它有用:

class A(object):

    def __new__(cls, children, *args, **kwargs):
        self = super(A, cls).__new__(cls)
        self._frozen = False  # allow mutation from here to end of  __init__
        # other stuff you need to do in __new__ goes here
        return self

    def __init__(self, *args, **kwargs):
        super(A, self).__init__()
        self._frozen = True  # prevent future mutation

    def __setattr__(self, name, value):
        # need to special case setting _frozen.
        if name != '_frozen' and self._frozen:
            raise TypeError('Instances are immutable.')
        else:
            super(A, self).__setattr__(name, value)

    def __delattr__(self, name):
        if self._frozen:
            raise TypeError('Instances are immutable.')
        else:
            super(A, self).__delattr__(name)

你可能需要根据用例重写更多的东西(比如__setitem__)。

其他回答

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)

..如何在C中“正确地”做这件事?

你可以使用Cython为Python创建一个扩展类型:

cdef class Immutable:
    cdef readonly object a, b
    cdef object __weakref__ # enable weak referencing support

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

它既适用于Python 2。X和3。

测试

# compile on-the-fly
import pyximport; pyximport.install() # $ pip install cython
from immutable import Immutable

o = Immutable(1, 2)
assert o.a == 1, str(o.a)
assert o.b == 2

try: o.a = 3
except AttributeError:
    pass
else:
    assert 0, 'attribute must be readonly'

try: o[1]
except TypeError:
    pass
else:
    assert 0, 'indexing must not be supported'

try: o.c = 1
except AttributeError:
    pass
else:
    assert 0, 'no new attributes are allowed'

o = Immutable('a', [])
assert o.a == 'a'
assert o.b == []

o.b.append(3) # attribute may contain mutable object
assert o.b == [3]

try: o.c
except AttributeError:
    pass
else:
    assert 0, 'no c attribute'

o = Immutable(b=3,a=1)
assert o.a == 1 and o.b == 3

try: del o.b
except AttributeError:
    pass
else:
    assert 0, "can't delete attribute"

d = dict(b=3, a=1)
o = Immutable(**d)
assert o.a == d['a'] and o.b == d['b']

o = Immutable(1,b=3)
assert o.a == 1 and o.b == 3

try: object.__setattr__(o, 'a', 1)
except AttributeError:
    pass
else:
    assert 0, 'attributes are readonly'

try: object.__setattr__(o, 'c', 1)
except AttributeError:
    pass
else:
    assert 0, 'no new attributes'

try: Immutable(1,c=3)
except TypeError:
    pass
else:
    assert 0, 'accept only a,b keywords'

for kwd in [dict(a=1), dict(b=2)]:
    try: Immutable(**kwd)
    except TypeError:
        pass
    else:
        assert 0, 'Immutable requires exactly 2 arguments'

如果你不介意索引支持,那么@Sven Marnach建议的collections.namedtuple是更可取的:

Immutable = collections.namedtuple("Immutable", "a b")

你可以创建一个@immutable装饰器,它覆盖__setattr__并将__slots__更改为一个空列表,然后用它装饰__init__方法。

编辑:正如OP所指出的,改变__slots__属性只会阻止新属性的创建,而不会阻止修改。

Edit2:下面是一个实现:

Edit3:使用__slots__会破坏这段代码,因为if会停止对象__dict__的创建。我正在寻找替代方案。

Edit4:嗯,就是这样。这是一个很粗鄙的问题,但可以作为练习:-)

class immutable(object):
    def __init__(self, immutable_params):
        self.immutable_params = immutable_params

    def __call__(self, new):
        params = self.immutable_params

        def __set_if_unset__(self, name, value):
            if name in self.__dict__:
                raise Exception("Attribute %s has already been set" % name)

            if not name in params:
                raise Exception("Cannot create atribute %s" % name)

            self.__dict__[name] = value;

        def __new__(cls, *args, **kws):
            cls.__setattr__ = __set_if_unset__

            return super(cls.__class__, cls).__new__(cls, *args, **kws)

        return __new__

class Point(object):
    @immutable(['x', 'y'])
    def __new__(): pass

    def __init__(self, x, y):
        self.x = x
        self.y = y

p = Point(1, 2) 
p.x = 3 # Exception: Attribute x has already been set
p.z = 4 # Exception: Cannot create atribute z

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__()。