Question about operator precedence for in and !=
Question:
While writing Python code, I got a result different from what I wanted.
>>> temp = [1]
>>> 1 in temp != 2 in temp
False
>>> (1 in temp) != (2 in temp)
True
>>> ((1 in temp) != 2) in temp
True
My purpose was the second, but I wrote it like the first.
The problem has been solved, but I wonder in what order the first expression outputs False.
I wondered if it was because of the same principle as the third, but the third also outputs True.
Answers:
I believe this is caused by operator chaining.
In Python, you can write an expression with two (or more) operators like this:
a < b < c
And Python treats this as if you wrote (a < b) and (b < c).
So Python is treating your expression
1 in temp != 2 in temp
As if you wrote:
(1 in temp) and (temp != 2) and (2 in temp)
Which is false.
While writing Python code, I got a result different from what I wanted.
>>> temp = [1]
>>> 1 in temp != 2 in temp
False
>>> (1 in temp) != (2 in temp)
True
>>> ((1 in temp) != 2) in temp
True
My purpose was the second, but I wrote it like the first.
The problem has been solved, but I wonder in what order the first expression outputs False.
I wondered if it was because of the same principle as the third, but the third also outputs True.
I believe this is caused by operator chaining.
In Python, you can write an expression with two (or more) operators like this:
a < b < c
And Python treats this as if you wrote (a < b) and (b < c).
So Python is treating your expression
1 in temp != 2 in temp
As if you wrote:
(1 in temp) and (temp != 2) and (2 in temp)
Which is false.