使用算术运算符将None与内置类型进行比较？

Question

Python 2.7.2 (default, Jun 12 2011, 14:24:46) [MSC v.1500 64 bit (AMD64)] on win32 
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>>> None > 0 
False 
>>> None == 0 
False 
>>> None < 0 
True 

• 是使用内置类型（整数在这种情况下）明确定义的算术运算符比较None？
• 语言规范的前两个和第三个比较部分之间的区别（Python的规范 - 你必须在开玩笑:)）还是CPython的实现细节？

Answer 1

您真正可以使用None的唯一有意义的比较是if obj is None:（或if obj is not None:）。

不同类型之间的比较已从Python 3中删除，原因很充分 - 它们是错误的常见来源，并导致混淆。例如

>>> "3" < 4 
False 

在Python 3，你比较不同类型的值时，像str与int或任何与None得到TypeError。

>>> None < 0 
Traceback (most recent call last): 
    File "<stdin>", line 1, in <module> 
TypeError: unorderable types: NoneType() < int() 

（我的意思是在试图确定哪两个数值越大/小平等比较正常意义上的“比较” - 它总是返回False如果两个对象是不同类型的。）

我没有找到的文档此的引用，但在Learning Python, 4th edition，马克·鲁茨204页上写道：

[...] Comparisons of differently typed objects (e.g., a string and a list) work — the language defines a fixed ordering among different types, which is deterministic, if not aesthetically pleasing. That is, the ordering is based on the names of the types involved: all integers are less than all strings, for example, because "int" is less than "str" .

Answer 2

一些有趣的报价从http://bytes.com/topic/python/answers/801701-why-none-0-a

In early Python, the decision was made that the comparison of any two objects was legal and would return a consistent result. So objects of different types will compare according to an ordering on their types (an implementation dependent, unspecified, but consistent ordering), and objects of the same type will be compared according to rules that make sense for that type.

Other implementations have the right to compare an integer and None differently, but on a specific implementation, the result will not change.

Python 3 will raise an exception on such comparisons.

和

The problem is the typical one; Python did not originally have a Boolean type, and the retrofit resulted in weird semantics. C has the same issue.