在 Python (2.7) 中比较两个相同的对象返回 False

2022-01-25 00:00:00 python python-2.7 compare sqlite

我在 Python 中有一个名为 object_from_DB 的函数.定义并不重要，只是它接受一个 ID 值作为参数，使用 sqlite3 库从 .db 文件中的表中提取匹配值，然后使用这些值作为参数对象的初始化.使用此功能不会更改数据库.

I have a function in Python called object_from_DB. The definition isn't important except that it takes an ID value as an argument, uses the sqlite3 library to pull matching values from a table in a .db file, and then uses those values as arguments in the initialization of an object. The database is in no way changed by the use of this function.

鉴于此，这个示例代码让我感到困惑.

This sample code, in light of this, baffles me. 

>>> x = object_from_DB(422)
>>> y = object_from_DB(422)
>>> x == y
False

为什么会发生这种情况，什么技术会导致 x 和 y 在比较时返回 True?

Why does this happen, and what sort of technique will cause x and y to return True when compared?

推荐答案

默认情况下，任何用户定义类的两个不同实例都是不相等的:

By default, two distinct instances of any user-defined class are unequal:

>>> class X: pass
... 
>>> a = X()
>>> b = X()
>>> a == b
False

如果你想要不同的行为，你必须定义它:

If you want different behaviour, you have to define it:

class Y:

    def __init__(self, value):
        self.value = value

    def __eq__(self, other):
        return self.value == other.value


>>> c = Y(3)
>>> d = Y(3)
>>> e = Y(4)
>>> c == d
True
>>> d == e
False

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