python - Comparing None with built-in types using arithmetic operators?

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

• Is comparing None using arithmetic operators well defined for built-in types (integers in this case)?
• Is the difference between the first two and the third comparison part of language specification (Python's specification - you must be kidding :)) or is it CPython's implementation detail?

Answer 1

The only meaningful comparison you can really use with None is if obj is None: (or if obj is not None:).

Comparison between different types has been removed from Python 3 for good reasons - they were a common source of errors and lead to confusion. For example

>>> "3" < 4
False

In Python 3, you get a TypeError when comparing values of different types like str vs. int or anything vs. None.

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

(I mean "comparing" in the sense of trying to determine which of two values is larger/smaller. Comparison for equality is OK - it will always return False if two object are of different types.)

I haven't found a reference in the docs for this, but in Learning Python, 4th edition, Mark Lutz writes on page 204:

[...] 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".