如何使类的特定实例不可变？

Question

拥有类 C 的实例 c，我想让 c 不可变，但 C 的其他实例不必这样做。

有没有一种简单的方法可以在 python 中实现这一点？

Answer 1

你不能让 Python 类完全不可变。不过你可以模仿它：

class C:
    _immutable = False
    def __setattr__(self, name, value):
        if self._immutable:
            raise TypeError(f"Can't set attribute, {self!r} is immutable.")
        super().__setattr__(name, value)

例子：

>>> c = C()
>>> c.hello = 123
>>> c.hello
123
>>> c._immutable = True
>>> c.hello = 456
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
  File "<stdin>", line 5, in __setattr__
TypeError: Can't set attribute, <__main__.C object at 0x000002087C679D20> is immutable.

如果你想在初始化时设置它，你可以添加__init__如下：

class C:
    _immutable = False
    def __init__(self, immutable=False):
        self._immutable = immutable
    def __setattr__(self, name, value):
        if self._immutable:
            raise TypeError(f"Can't set attribute, {self!r} is immutable.")
        super().__setattr__(name, value)

请记住，您仍然可以通过__dict__直接访问和修改实例来绕过它：

>>> c = C(immutable=True)
>>> c.__dict__["hello"] = 123
>>> c.hello
123

您可以尝试像这样阻止它：

class C:
    _immutable = False
    def __init__(self, immutable=False):
        self._immutable = immutable
    def __getattribute__(self, name):
        if name == "__dict__":
            raise TypeError("Can't access class dict.")
        return super().__getattribute__(name)
    def __setattr__(self, name, value):
        if self._immutable:
            raise TypeError(f"Can't set attribute, {self!r} is immutable.")
        super().__setattr__(name, value)

但即便如此，也可以绕过：

>>> c = C(immutable=True)
>>> object.__getattribute__(c, "__dict__")["hello"] = 123
>>> c.hello
123