How does Python parse the first of the following three expressions? (I expected it to be the same as the second, since == and in have the same precedence.)
>>> 1 == 0 in (0,1), (1==0) in (0,1), 1 == (0 in (0,1))
(False, True, True)
Asked
Active
Viewed 126 times
5
Alan
- 9,410
- 15
- 20
-
The confusion is a common one and finding the canonical is always going to be hard as the search terms involve punctuation that most search engines ignore. Your post can help provide additional keywords to guide people to the canonical, however. I'd keep it. – Martijn Pieters May 01 '15 at 22:59
1 Answers
5
See the documentation of comparison operators: they are chained rather than grouped. So 1 == 0 in (0,1) is equivalent to (1==0) and (0 in (0,1)), which is obviously false.
Alan
- 9,410
- 15
- 20
-
sort of ... its more equivelent to `False in (0,1)` but still true – Joran Beasley May 01 '15 at 21:22
-
@JoranBeasley Incorrect. `False in (0,1)` yields `True`, as shown in the question. – John Kugelman May 01 '15 at 21:25
-
1OK OK you win I was wrong... I didnt read the answer close enough :P – Joran Beasley May 01 '15 at 21:29