Why does Python evaluate "foo" in "bar" != "baz" as False?

Asked Apr 22 '21 at 18:46

Active Apr 22 '21 at 18:46

Viewed 34 times

While reviewing some Python code, I encountered some bad logic. But, I don't understand how it is getting evaluated. The code amounts to this:

>>> "foo" in "bar" != "baz"
False

How is this False? The "in" and "!=" have the same order of precedence. If the "in" is evaluated first:

>>> ("foo" in "bar") != "baz"
True

because False is not equal to "baz". But if "!=" has precedence, then this should raise an exception:

>>> "foo" in ("bar" != "baz")
Traceback (most recent call last):
  File "<pyshell#8>", line 1, in <module>
    "foo" in ("bar" != "baz")
TypeError: argument of type 'bool' is not iterable

So, for the first example, without parentheses, what is going on?

asked Apr 22 '21 at 18:46

Andy Vancil

TL;DR `"foo" in "bar" != "baz"` is evaluated as `"foo" in "bar" and "bar" != "baz"` – DeepSpace Apr 22 '21 at 18:50
1

Also, use parenthesis. There's no awards for having less characters in the source file. – Donnie Apr 22 '21 at 18:51
Thanks. So, to summarize, this is treated as a chained operation. Not intuitive, but something to keep in mind. – Andy Vancil Apr 23 '21 at 19:23

Why does Python evaluate "foo" in "bar" != "baz" as False?

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