这个python语法发生了什么？(c == c in s)

Question

有人向我展示了这个奇怪的python语法示例.为什么[4]有效？

我原以为它要评估[5]或[6],两者都不起作用.是否有一些不成熟的过早优化？

In [1]: s = 'abcd'

In [2]: c = 'b'

In [3]: c in s
 Out[3]: True

In [4]: c == c in s
Out[4]: True

In [5]: True in s
---------------------------------------------------------------------------
TypeError                                 Traceback (most recent call last)
<ipython-input-5-e00149345694> in <module>()
----> 1 True in s

TypeError: 'in <string>' requires string as left operand, not bool

In [6]: c == True
Out[6]: False

Answer 1

这与允许python将多个运算符(如<)链接在一起的语法糖相同.

例如:

>>> 0 < 1 < 2
True

这相当于(0<1) and (1<2),只有中间表达式只被评估一次.

该陈述c == c in s同样等同于(c == c) and (c in s),其评估结果为True.

要突出显示前一点,中间表达式仅评估一次:

>>> def foo(x):
...     print "Called foo(%d)" % x
...     return x
...
>>> print 0 < foo(1) < 2
Called foo(1)
True

有关更多详细信息,请参阅Python语言参考.