How is this expression evaluated? "f" in "foo" == True

Question

>>> "f" in "foo"
True
>>> "f" in "foo" == True
False

I'm confused why the second expression is False. I see == has higher precedence than in. But then I would expect to get an exception, which is what happens when I add parentheses:

>>> "f" in ("foo" == True)
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
TypeError: argument of type 'bool' is not iterable

It seems the expression is only True when foo is on both sides of ==, like so:

>>> "f" in "foo" == "foo"
True
>>> "f" in "foo" == "bar"
False

What am I missing? What is Python actually calculating here?

"I see == has higher precedence than in" - `in` and `==` have equal precedence. — sj95126, Oct 23 '21 at 21:44

score 9 · Accepted Answer · answered Oct 23 '21 at 21:43

In Python, comparison operators chain.

That is why 1 < 2 < 3 < 4 works and evaluates to True.

See https://docs.python.org/3/reference/expressions.html#comparisons

in and == are such operators, so by this mechanism

"f" in "foo" == True

means

("f" in "foo") and ("foo" == True)

How is this expression evaluated? "f" in "foo" == True

1 Answers1