Python NotImplemented 常量

Python NotImplemented constant

提问by Anurag Uniyal

Looking through decimal.py, it uses NotImplementedin many special methods. e.g.

翻看decimal.py，它用NotImplemented在许多特殊的方法中。例如

class A(object):
    def __lt__(self, a):
        return NotImplemented

    def __add__(self, a):
        return NotImplemented

The Python docs say:

在Python的文件说：

NotImplemented

Special value which can be returned by the “rich comparison” special methods (__eq__(), __lt__(), and friends), to indicate that the comparison is not implemented with respect to the other type.

未实现

可以通过“丰富比较”特殊方法（__eq__()、__lt__()、 和朋友）返回的特殊值，以指示未针对其他类型进行比较。

It doesn't talk about other special methods and neither does it describe the behavior.

它没有讨论其他特殊方法，也没有描述行为。

It seems to be a magic object which if returned from other special methods raises TypeError, and in “rich comparison” special methods does nothing.

它似乎是一个魔术对象，如果从其他特殊方法返回，则会引发TypeError，而在“丰富的比较”中，特殊方法什么也不做。

e.g.

例如

print A() < A()

prints True, but

打印True，但是

print A() + 1

raises TypeError, so I am curious as to what's going on and what is the usage/behavior of NotImplemented.

raises TypeError，所以我很好奇发生了什么以及 NotImplemented 的用法/行为是什么。

回答by Brandon E Taylor

NotImplementedallows you to indicate that a comparison between the two given operands has not been implemented (rather than indicating that the comparison is valid, but yields False, for the two operands).

NotImplemented允许您指示尚未实现两个给定操作数之间的比较（而不是指示比较有效，但False对两个操作数产生 ）。

From the Python Language Reference:

来自Python 语言参考：

For objects x and y, first x.__op__(y)is tried. If this is not implemented or returns NotImplemented, y.__rop__(x)is tried. If this is also not implemented or returns NotImplemented, a TypeError exception is raised. But see the following exception:

Exception to the previous item: if the left operand is an instance of a built-in type or a new-style class, and the right operand is an instance of a proper subclass of that type or class and overrides the base's __rop__()method, the right operand's __rop__()method is tried before the left operand's __op__()method. This is done so that a subclass can completely override binary operators. Otherwise, the left operand's __op__()method would always accept the right operand: when an instance of a given class is expected, an instance of a subclass of that class is always acceptable.

对于对象 x 和 y，首先x.__op__(y)尝试。如果这未实现或返回 NotImplemented， y.__rop__(x)则尝试。如果这也未实现或返回 NotImplemented，则会引发 TypeError 异常。但看到以下异常：

上一项的例外：如果左操作数是内置类型或新式类的实例，而右操作数是该类型或类的适当子类的实例并覆盖基类的__rop__()方法，则右操作数操作的__rop__()方法是左操作的尝试过__op__()的方法。这样做是为了让子类可以完全覆盖二元运算符。否则，左操作数的__op__()方法将始终接受右操作数：当需要给定类的实例时，该类的子类的实例总是可接受的。

回答by Anurag Uniyal

It actually has the same meaning when returned from __add__as from __lt__, the difference is Python 2.x is trying other ways of comparing the objects before giving up. Python 3.x does raise a TypeError. In fact, Python can try other things for __add__as well, look at __radd__and (though I'm fuzzy on it) __coerce__.

当从__add__from返回时它实际上具有相同的含义__lt__，不同之处在于 Python 2.x 在放弃之前尝试其他比较对象的方法。Python 3.x 确实会引发 TypeError。事实上，Python 也可以尝试其他的东西__add__，看看__radd__和（虽然我对它很模糊）__coerce__。

# 2.6
>>> class A(object):
...   def __lt__(self, other):
...     return NotImplemented
>>> A() < A()
True

# 3.1
>>> class A(object):
...   def __lt__(self, other):
...     return NotImplemented
>>> A() < A()
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
TypeError: unorderable types: A() < A()

See Ordering Comparisions (3.0 docs)for more info.

有关更多信息，请参阅订购比较（3.0 文档）。

回答by Lennart Regebro

If you return it from __add__it will behave like the object has no __add__method, and raise a TypeError.

如果您从中返回它，__add__它将表现得像对象没有__add__方法一样，并引发一个TypeError.

If you return NotImplementedfrom a rich comparison function, Python will behave like the method wasn't implemented, that is, it will defer to using __cmp__.

如果NotImplemented从富比较函数返回，Python 将表现得好像该方法未实现，也就是说，它将推迟使用__cmp__.