除了琐碎的数据类型，python 不可变对象 ID 是否相同？

Question

对于不可变数据类型，Python不保证两个构造产生两个不同的对象；因此，1+1 is 2 可能会或可能不会 return True。当然，mutable 类型 do 会产生单独的对象：[] is [] 将始终 return False.

来自 https://docs.python.org/3/reference/datamodel.html#objects-values-and-types:

Types affect almost all aspects of object behavior. Even the importance of object identity is affected in some sense: for immutable types, operations that compute new values may actually return a reference to any existing object with the same type and value, while for mutable objects this is not allowed. E.g., after a = 1; b = 1, a and b may or may not refer to the same object with the value one, depending on the implementation, but after c = []; d = [], c and d are guaranteed to refer to two different, unique, newly created empty lists. (Note that c = d = [] assigns the same object to both c and d.)

>>> "abc" is "abc"
True
>>> [] is []
False
>>> 1 + 1 is 2
True

现在，除了琐碎的类型 int、float、bool、string，我希望 tuple 和 frozenset表现出相同的行为。

但我没有发现他们这样做的情况：

>>> {1, 2} is {1, 2}
False
>>> (1, ) is (1, )
False

是否有任何实现为单独的不可变对象产生相同的 id？

Answer 1

这些对象的所有 is 相等性都是解释器技巧的结果（这不是语言规范 AFAIK 的一部分，而是 CPython 的结果）

例如(with int).

The current implementation keeps an array of integer objects for all integers between -5 and 256, when you create an int in that range you actually just get back a reference to the existing object.

所以应该可以更改 1 的值。我怀疑 Python 在这种情况下的行为未定义。 :-)

对于字符串，python 会 interning。 bool 是 int (issubclass(bool, int)) 的子类，它解释了 bool 的共享 id。 AFAIK，浮动（除了 3.8 中的文字）永远不会 return True for is equality.

在 python3.8 中，(1,) is (1,) 是 actually True，因为文字计算发生了变化。

>>> (1,) is (1,)
<stdin>:1: SyntaxWarning: "is" with a literal. Did you mean "=="?
True

连同 is 用于文字的警告。但是，分配变量和执行比较是错误的（如预期的那样）。

请注意 tuple() is tuple() 和 frozenset() is frozenset() 都 return 正确。但是，它们的可变对象（list 和 set）没有这个属性.

除了琐碎的数据类型，python 不可变对象 ID 是否相同？

Apart from trivial data types, then are python immutable object ids the same?

python

optimization

inequality

immutability