"In" Check not working in if Statement

Question

I was doing some basic string manipulation and I came across something very interesting and confusing:

str= "HELLO WORLD"
x="LL"

t = x in str
print t

if x in str == True:
    print "TRUE"

When it's run, it currently only prints True on one occasion, on the first print statement. One would think that both would print as logically and almost syntactically they are the same yet one does not. I do not understand the reasoning behind this, especially given how Python is supposed to be intuitive.

FWIW, the `== True` is entirely redundant and can/should be omitted. — deceze, Dec 22 '16 at 13:02
use without `== True` - `if x in str:` - now it is treated as `if x in (str == True):` — furas, Dec 22 '16 at 13:03
@furas: No, if it were parsed that way, there'd be a `TypeError`. The `in` and `==` actually apply comparison chaining, like in `1 < x < 5`. — user2357112, Dec 22 '16 at 13:05
and `str` is a built-in; you should never overwrite any of those or you will encounter unpleasant surprises... — hiro protagonist, Dec 22 '16 at 13:06
Operator chaining should really be limited to ==, <=, !=, >= in my opinion, this sort of thing is *always* surprising. But too late for that. — RemcoGerlich, Dec 22 '16 at 13:29

Jean-François Fabre · Accepted Answer · 2016-12-22T13:07:45.300

Protect your condition into parentheses:

if (x in str) == True:
    print("TRUE")

Even better: it's redundant to compare to True

if x in str:
    print("TRUE")

That said, it's not beacuse of operator precedence:

x in (str == True)

is invalid (cannot iterate on a boolean)

x in str == True

is valid but returns False. I must admit I'm still puzzled about that...

EDIT: just saw the duplicate and now I get it. Lesson #1: always protect your expressions with parentheses when you have a doubt.

"In" Check not working in if Statement

1 Answers1