为什么string>int的计算结果为True？

Question

检查字符串int>；如何实现？你知道吗

>>> strver = "1"
>>> ver = 1
>>> strver > ver
True
>>> strVer2 = "whaat"
>>> strVer2 > ver
True

做了更多的实验：

>>> ver3 = 0
>>> strVer2 > ver3
True

我认为在尝试比较时应该有一个错误，但似乎没有生成任何东西来处理这样的错误，或者应该使用assert，但是如果python代码是用-O标志运行的，那么这可能是危险的！你知道吗

Answer 1

来源：How does Python compare string and int?，它反过来引用了CPythonmanual：

CPython implementation detail: Objects of different types except numbers are ordered by their type names; objects of the same types that don’t support proper comparison are ordered by their address.

从答案来看：

When you order two incompatible types where neither is numeric, they are ordered by the alphabetical order of their typenames:

>>> [1, 2] > 'foo'   # 'list' < 'str' 
False
>>> (1, 2) > 'foo'   # 'tuple' > 'str'
True

>>> class Foo(object): pass
>>> class Bar(object): pass
>>> Bar() < Foo()
True

……所以，这是因为字母表中“s”在“i”之后！不过，幸运的是，在Python 3.x的实现中，这个稍微奇怪的行为已经被“修复”：

In Python 3.x the behaviour has been changed so that attempting to order an integer and a string will raise an error:

现在似乎遵循最小惊奇的原则好一点了。你知道吗