如何在两个不同的内存位置创建int 1？ [英] How to create the int 1 at two different memory locations?

问题描述

我想向某人展示如何使用 is 而不是 == 比较整数会失败。我以为这可以用，但是没有用。

I want to show someone how using is instead of == to compare integers can fail. I thought this would work, but it didn't:

>>> import copy
>>> x = 1
>>> y = copy.deepcopy(x)
>>> x is y
True

对于较大的整数，我可以轻松做到这一点：

I can do this easily for bigger integers:

>>> x = 500
>>> y = 500
>>> x is y
False

我如何用较小的整数演示同一件事？

How can I demonstrate the same thing with smaller integers which might typically be used for enum-like purposes in python?

推荐答案

以下示例在Python 2和3中均失败：

The following example fails in both Python 2 and 3:

>>> n=12345
>>> ((n**8)+1) % (n**4) is 1
False
>>> ((n**8)+1) % (n**4) == 1
True

原因略有不同。 Python 2对小整数使用 int 类型，对任意精度值使用 long 类型。仅实习 int 类型，因此当返回 1L 时该示例失败。

The reasons are slightly different. Python 2 uses the int type for small integers and the long type for arbitrary precision values. Only the int type is interned so the example fails when a 1L is returned.

Python 3仅使用任意精度类型（并将其重命名为 int ）。该示例失败，因为余数计算在内部计算值1并将其返回。仅当创建对象并且在计算开始时创建对象且该对象的值为1之前，才进行内部检查。

Python 3 only uses the arbitrary precision type (and renamed it to int). The example fails because the remainder calculation internally computes a value of 1 and returns it. The interning check is only done when objects are created and the object was created at the start of the calculation before it had the value 1.

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